Solution

Number of letters in the word ‘ALGORITHM’ $=9$

If ‘GOR’ remain together, then considered it as 1 group.

$\therefore$ Number of letters $=6+1=7$

Number of word, if ‘GOR’ remain together $=7$ !

Total number of words from the letters of the word ‘ALGORITHM’ $=9$ !

$\therefore \quad$ Required probability $=\frac{7 !}{9 !}=\frac{1}{72}$

Solution

Let the couple occupied adjacent desks consider those two as 1 .

There are $(4+1)$ i.e ., 5 persons to be assigned.

$\therefore \quad$ Number of ways of assigning these five person $=5$ ! $\times 2$ !

Total number of ways of assigning 6 persons $=6$ !

$\therefore \quad$ Probability that the couple has adjacent desk $=\frac{5 ! \times 2 !}{6 !}=\frac{2}{6}=\frac{1}{3}$

Probability that the married couple will have non-adjacent desks $=1-\frac{1}{3}=\frac{2}{3}$

Solution

Multiple of 2 from 1 to 1000 are 2, 4, 6, 8, .., 1000 Let $n$ be the number of terms of above series.

$\therefore \quad n$ th term $=1000$

$\Rightarrow \quad 2+(n-1) 2=1000$

$\Rightarrow \quad 2+2 n-2=1000$

$\Rightarrow \quad 2 n=1000$

$\therefore \quad n=500$

Since, the number of multiple of 2 are 500 .

So, the multiple of 9 are $9,18,27, \ldots, 999$

Let $m$ be the number of term in above series.

$\therefore \quad m$ th term $=999$

$\Rightarrow \quad 9+(m-1) 9=999$

$\Rightarrow \quad 9+9 m-9=999$

$\Rightarrow \quad 9 m=999$

$\therefore \quad m=111$

Since, the number of multiple of 9 are 111 . So, the multiple of 2 and 9 both are $18,36, \ldots$, 990

Let $p$ be the number of terms in above series.

$\therefore \quad$ pth term $=990$

$\Rightarrow \quad 18+(p-1) 18=990$

$\Rightarrow \quad 18+18 p-18=990$

$\Rightarrow \quad 18 p=990$

$\therefore \quad p=\frac{990}{18}=55$

Since, the number of multiple of 2 and 9 are 55 .

$\therefore \quad$ Number of multiple of 2 or $9=500+111-55=556$

$\therefore$ Required probability $=\frac{n(E)}{n(S)}=\frac{556}{1000}=0.556$

Solution

In a through of a die there is 6 sample points.

(i) If 2 appears on the $k$ th roll of the die.

So, first $(k-1)$ roll have 5 outcomes each and $k$ th roll results 2 i.e., 1 outcome.

$\therefore$ Number of element of sample space correspond to the event that 2 appears on the $k$ th roll of the die $=5^{k-1}$

(ii) If we consider that 2 appears not later than $k$ th roll of the die, then it is possible that 2 comes in first throw i.e., 1 outcome.

If 2 does not appear in first throw, then outcomes will be 5 and 2 comes in second throw i.e., 1 outcome, possible outcome $=5 \times 1=5$

Similarly, if 2 does not appear in second throw and appears in third throw.

$\therefore \quad$ Possible outcomes $=5 \times 5 \times 1$

Given,

$ \begin{aligned} \text { series } & =1+5+5 \times 5+5 \times 5 \times 5+\ldots+5^{k-1} \\ & =1+5+5^{2}+5^{3}+\ldots+5^{k-1} \\ & =\frac{1(5^{k}-1)}{5-1}=\frac{5^{k}-1}{4} \end{aligned} $

Solution

It is given that, $2 \times$ Probability of even number $=$ Probability of odd number

$\Rightarrow$ $ P(O)=2 P(E) $

$\Rightarrow$ $ P(O): P(E)=2: 1 $

$\therefore$ Probability of occurring odd number, $P(O)=\frac{2}{2+1}=\frac{2}{3}$

and probability of occurring 5each number,

$ P(E)=\frac{1}{2+1}=\frac{1}{3} $

Now, $G$ be the even that a number greater than 3 occur in a single roll of die.

So, the possible outcomes are 4,5 and 6 out of which two are even and one odd.

$\therefore \quad$ Required probability $=P(G)=2 \times P(E) \times P(O)$

$ =2 \times \frac{1}{3} \times \frac{2}{3}=\frac{4}{9} $

Solution

Let $E_1$ be the event that family own colour television set and $E_2$ be the event that family owns a black and white television set.

It is given that,

$P(E_1)=0.87$

$ P(E_2)=0.36 $

and $ P(E_1 \cap E_2)=0.30 $

We have to find probability that a family owns either anyone or both kind of sets i.e., $P(E_1 \cup E_2)$.

Now, $ \begin{aligned} P(E_1 \cup E_2) & =P(E_1)+P(E_2)-P(E_1 \cap E_2) \quad \text { [by addition theorem] } \\ & =0.87+0.36-0.30 \\ & =0.93 \end{aligned} $

Solution

Since, it is given that, $A$ and $B$ are mutually exclusive events. $\therefore$ $P(A \cap B)=0$

and $P(A)=0.35, P(B)=0.45$

$[\because A \cap B=\varphi]$

(i) $P(A^{\prime})=1-P(A)=1-0.35=0.65$

(ii) $P(B^{\prime})=1-P(B)=1-0.45=0.55$

(iii) $P(A \cup B)=P(A)+P(B)-P(A \cap B)=0.35+0.45-0=0.80$

(iv) $P(A \cap B)=0$

(v) $P(A \cap B^{\prime})=P(A)-P(A \cap B)=0.35-0=0.35$

(vi) $P(A^{\prime} \cap B^{\prime})=P(A \cup B)^{\prime}=1-P(A \cup B)=1-0.8=0.2$

Solution

Let $E_1, E_2, E_3, E_4$ and $E_5$ be the event that surgeries are rated as very complex, complex, routine, simple or very simple, respectively.

$\therefore \quad P(E_1)=0.15, P(E_2)=0.20, P(E_3)=0.31, P(E_4)=0.26, P(E_5)=0.08$

(i) $P$ (complex or very complex) $=P(E_1.$ or $.E_2)$

$ \begin{aligned} & =P(E_1 \cup E_2)=P(E_1)+P(E_2)-P(E_1 \cap E_2) \\ & =0.15+0.20-0[P(E_1 \cap E_2)=0. \\ & \quad \quad \text { because all events are independent }] \\ & =0.35 \quad \end{aligned} $

(ii) $P$ (neither very complex nor very simple), $(P(E_1^{\prime} \cap E_5^{\prime})=P(E_1 \cup E_5)^{\prime}.$

$ \begin{aligned} & =1-P(E_1 \cup E_5) \\ & =1-[P(E_1)+P(E_5)] \\ & =1-(0.15+0.08) \\ & =1-0.23 \\ & =0.77 \end{aligned} $

(iii) $P$ (routine or complex) $=P(E_3 \cup E_2)=P(E_3)+P(E_2)$

$ =0.31+0.20=0.51 $

(iv) $P$ (routine or simple) $=P(E_3 \cup E_4)=P(E_3)+P(E_4)$

$ =0.31+0.26=0.57 $

Solution

It is given that $A$ is twice as likely to be selected as $D$.

$ \begin{aligned} & P(A)=2 P(B) \\ & \frac{P(A)}{2}=P(B) \end{aligned} $

while $C$ is twice as likely to be selected as $D$.

$ \begin{aligned} P(C) =2 P(D) \Rightarrow \quad P(B)=2 P(D) \\ \Rightarrow \frac{P(A)}{2} =2 P(D) \Rightarrow \quad P(D)=\frac{P(A)}{4} \end{aligned} $

$B$ and $C$ are given about the same chance of being selected

$ P(B)=P(C) $

Now,

$ \text { sum of probability }=1 $

$ \begin{aligned} P(A)+P(B)+P(C)+P(D) =1 \\ P(A)+\frac{P(A)}{2}+P \frac{(A)}{2}+\frac{P(A)}{4} =1 \\ \Rightarrow \quad \frac{4 P(A)+2 P(A)+2 P(A)+P(A)}{4} =1 \\ \Rightarrow \quad 9 P(A) =4 \Rightarrow P(A)=\frac{4}{9} \\ \text { (i) } P(C \text { will be selected })=P(C)=P(B)= & \frac{P(A)}{2} \\ & =\frac{4}{9 \times 2} \\ & =\frac{2}{9} \end{aligned} $

(ii) $P(A$ will not be selected $)=P(A^{\prime})=1-P(A)=1-\frac{4}{9}=\frac{5}{9}$

Solution

Let

$ \begin{aligned} E_1 & =\text { John promoted } \\ E_2 & =\text { Rita promoted } \\ E_3 & =\text { Aslam promoted } \\ E_4 & =\text { Gurpreet promoted } \end{aligned} $

Given, sample space, $S=\lbrace$ John promoted, Rita promoted, Aslam promoted, Gurpreet promoted $\rbrace$

i.e., $\quad S=\lbrace E_1, E_2, E_3, E_4\rbrace$

It is given that, chances of John’s promotion is same as that of Gurpreet.

$ P(E_1)=P(E_4) $

Rita’s chances of promotion are twice as likely as John.

$ P(E_2)=2 P(E_1) $

And Aslam’s chances of promotion are four times that of John.

$ P(E_3)=4 P(E_1) $

Now, $\quad P(E_1)+P(E_2)+P(E_3)+P(E_4)=1$

$\Rightarrow \quad P(E_1)+2 P(E_1)+4 P(E_1)+P(E_1)=1$

$\Rightarrow \quad 8 P(E_1)=1$

$\therefore \quad P(E_1)=\frac{1}{8}$

(i) $P($ John promoted $)=P(E_1)=\frac{1}{8}$

$P($ Rita promoted $)=P(E_2)=2 P(E_1)=2 \times \frac{1}{8}=\frac{2}{8}=\frac{1}{4}$

$P($ Aslam promoted $)=P(E_3)=4 P(E_1)=4 \times \frac{1}{8}=\frac{1}{2}$

$P($ Gurpreet promoted $)=P(E_4)=P(E_1)=\frac{1}{8}$

(ii) $A=$ John promoted or Gurpreet promoted

$ \begin{aligned} A & =E_1 \cup E_4 \\ P(A)=P(E_1 \cup E_4) & =P(E_1)+P(E_4)-P(E_1 \cap E_4) \quad[\because P(E_1 \cap E_4)=0] \\ & =\frac{1}{8}+\frac{1}{8}-0 \\ & =\frac{2}{8}=\frac{1}{4} \end{aligned} $

Solution

From the above Venn diagram,

(i) $P(A)=0.13+0.07=0.20$

(ii) $P(B \cap \bar{C})=P(B)-P(B \cap C)=0.07+0.10+0.15-0.15=0.07+0.10=0.17$

(iii) $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$ \begin{aligned} & =0.13+0.07+0.07+0.10+0.15-0.07 \\ & =0.13+0.07+0.10+0.15=0.45 \end{aligned} $

(iv) $P(A \cap \bar{B})=P(A)-P(A \cap B)=0.13+0.07-0.07=0.13$

(v) $P(B \cap C)=0.15$

(vi) $P$ (exactly one of the three occurs) $=0.13+0.10+0.28=0.51$

Solution

It is given that one of the two urn is chosen, then a ball is randomly chosen from the urn, then a second ball is chosen at random from the same urn without replacing the first ball.

(i) $\therefore\text{Sample space} S=\lbrace B_1 B_2, B_1 W, B_2 B_1, B_2 W, W B_1, W B_2, B W_1, B W_2, W_1 B, W_1 W_2, W_2 B, W_2 W_1\rbrace$

$\therefore$ total sample point $=12$

(ii) If two black ball are chosen.

So, the favourable events are $B_1 B_2, B_2 B_1$ i.e., 2

$\therefore$ Required probability $=\frac{2}{12}=\frac{1}{6}$

(iii) If two balls of opposite colour are chosen.

So, the favourable events are $B_1 W_1, B_2 W_1, W B_1, W B_2, B W_1, B W_2, W_1 B_1, W_2 B i . e ., 8$.

$\therefore$ Required probability $=\frac{8}{12}=\frac{2}{3}$

Solution

$\because$ Number of red balls $=8$

and number of white balls $=5$

(i) $P($ all the three balls are white $)=\frac{{ }^{5} C_3}{{ }^{13} C_3}=\frac{\frac{5 !}{3 ! 2 !}}{\frac{13 !}{3 ! 10 !}}=\frac{5 !}{3 ! 2 !} \times \frac{3 ! 10 !}{13 !}$

$ \begin{aligned} & =\frac{5 \times 4 \times 3 \times 2 !}{2 !} \times \frac{10 !}{13 \times 12 \times 11 \times 10 !}=\frac{5 \times 4 \times 3}{13 \times 12 \times 11} \\ & =\frac{5}{13 \times 11}=\frac{5}{143} \\ & =\frac{5 \times 4 \times 3}{13 \times 12 \times 11}=\frac{5}{13 \times 11}=\frac{5}{143} \end{aligned} $

(ii) $P$ (all the three balls are red)

$ \begin{aligned} & =\frac{{ }^{8} C_3}{{ }^{13} C_3}=\frac{\frac{8 !}{3 ! 5 !}}{\frac{13 !}{3 ! 10 !}}=\frac{8 !}{3 ! \times 5 !} \times \frac{3 ! 10 !}{13 !} \\ & =\frac{8 \times 7 \times 6 \times 5 !}{5 !} \times \frac{10 !}{13 \times 12 \times 11 \times 10 !} \\ & =\frac{8 \times 7 \times 6}{13 \times 12 \times 11}=\frac{28}{143} \end{aligned} $

(iii) $P$ (one ball is red and two balls are white)

$ =\frac{{ }^{8} C_1 \times{ }^{5} C_2}{{ }^{13} C_3}=\frac{8 \times 10}{13 \times 6 \times 11}=\frac{40}{143} $

Solution

Total number of letters in the word ‘ASSASSINATION’ are 13.

Out of which 3A’s, 4S’s, 2 l’s, 2 N’s, 1 T’s and 10.

(i) If four S’s come consecutively in the word, then we considers these 4 S’s as 1 group. Now, the number of laters is 10.

Number of words when all S’s are together $=\frac{10 !}{3 ! 2 ! 2 !}$

Total number of word using letters of the word ‘ASSASSINATION’

$ \begin{aligned} & =\frac{13 !}{3 ! 4 ! 2 ! 2 !} \\ \therefore \quad \text { Required probability } & =\frac{10 !}{\frac{3 ! 2 ! 2 ! \times 13 !}{3 ! 4 ! 2 ! 2 !}} \\ & =\frac{10 ! \times 4 !}{13 !}=\frac{4 !}{13 \times 12 \times 11}=\frac{24}{1716}=\frac{2}{143} \end{aligned} $

(ii) If 2 l’s and $2 N$ ’s come together, then there as 10 alphabets.

Number of word when $2 I$ ’s and $2 N$ ’s are come together

$ \begin{aligned} & =\frac{10 !}{3 ! 4 !} \times \frac{4 !}{2 ! 2 !} \\ \therefore \quad \text { Required probability } & =\frac{\frac{10 ! 4 !}{3 ! 4 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}=\frac{4 ! 10 !}{2 ! 2 ! 3 ! 4 !} \times \frac{3 ! 4 ! 2 ! 2 !}{13 !} \\ & =\frac{4 ! 10 !}{13 !}=\frac{4 !}{13 \times 12 \times 11}=\frac{24}{13 \times 12 \times 11}=\frac{2}{143} \end{aligned} $

(iii) If all A’s are coming together, then there are 11 alphabets.

Number of words when all A’s come together

$ =\frac{11 !}{4 ! 2 ! 2 !} $

Probability when all A’s come together

$ =\frac{\frac{11 !}{4 ! 2 ! 2 !}}{\frac{13 !}{4 ! 3 ! 2 ! 2 !}}=\frac{11 !}{4 ! 2 ! 2 !} \times \frac{4 ! 3 ! 2 ! 2 !}{13 !}=\frac{11 ! \times 3 !}{13 !}=\frac{6}{13 \times 12}=\frac{1}{26} $

Required probability when all A’s does not come together

$ =1-\frac{1}{26}=\frac{25}{26} $

(iv) If no two A’s are together, then first we arrange the alphabets except A’s.

All the alphabets except A’s are arranged in $\frac{10 !}{4 ! 2 ! 2 !}$.

There are 11 vacant places between these alphabets.

So, 3 A’s can be place in 11 places in ${ }^{11} C_3$ ways $=\frac{11 !}{3 ! 8 !}$

$\therefore \quad$ Total number of words when no two A’s together

$ \begin{aligned} & =\frac{11 !}{3 ! 8 !} \times \frac{10 !}{4 ! 2 ! 2 !} \\ \text { Required probability } & =\frac{11 ! \times 10 !}{3 ! 8 ! 4 ! 2 ! 2 !} \times \frac{4 ! 3 ! 2 ! 2 !}{13 !}=\frac{10 !}{8 ! \times 13 \times 12} \\ & =\frac{10 \times 9}{13 \times 12}=\frac{90}{156}=\frac{15}{26} \end{aligned} $

Solution

$\because$ Number of possible event $=52$

and favourable events $=4 king+13$ heart +26 red $-13-2=28$

$ \therefore \quad \text { Required probability }=\frac{28}{52}=\frac{7}{13} $

Solution

Given,

$ \begin{aligned} S & =\lbrace E_1, E_2, E_3, E_4, E_5, E_6, E_7, E_8, E_9\rbrace \\ A & =\lbrace E_1, E_5, E_8\rbrace, B=\lbrace E_2, E_5, E_8, E_9\rbrace \\ P(E_1) & =P(E_2)=0.08 \\ P(E_3) & =P(E_4)=P(E_5)=0.1. \\ P(E_6) & =P(E_7)=2, P(E_8)=P(E_9)=0.07 \end{aligned} $

(i) $P(A)=P(E_1)+P(E_5)+P(E_8)$

$ =0.08+0.1+0.07=0.25 $

(ii) $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

Now, $P(B)=P(E_2)+P(E_5)+P(E_8)+P(E_9)$

$ =0.08+0.1+0.07+0.07=0.32 $

$A \cup B=\lbrace E_5, E_8\rbrace$

$P(A \cap B)=P(E_5)+P(E_8)=0.1+0.7=0.17$

On substituting these values in E

(i), we get

$ P(A \cup B)=0.25+0.32-0.17=0.40 $

(iii) $A \cup B=\lbrace E_1, E_2, E_5, E_8, E_9\rbrace$

$ \begin{aligned} P(A \cup B) & =P(E_1)+P(E_2)+P(E_5)+P(E_8)+P(E_9) \\ & =0.08+0.08+0.1+0.07+0.07=0.40 \end{aligned} $

$ \text { (iv) } \because \begin{aligned} \because P(\bar{B})=1-P(B) & =1-0.32=0.68 \\ \text { and } \quad \bar{B} & =\lbrace E_1, E_3, E_4, E_6, E_7\rbrace \\ \therefore \quad P(\bar{B}) & =P(E_1)+P(E_3)+P(E_4)+P(E_6)+P(E_7) \\ & =0.08+0.1+0.1+0.2+0.2=0.68 \end{aligned} $

Solution

(i) When a die is throw the possible outcomes are

$S=\lbrace1,2,3,4,5,6\rbrace$ out of which $1,3,5$ are odd,

$\therefore$ Required probability $=\frac{3}{6}=\frac{1}{2}$

(ii) When a fair coin is tossed two times the sample space is

$ S=\lbrace H H, H T, T H, T T\rbrace $

In at least one head favourable enonts are $H H, H T, T H$

$\therefore$ Required probability $=\frac{3}{4}$

(iii) Total cards $=52$

$ \text { Favourable }=4 \text { king }+2 \text { of heart }+3 \text { of spade }=4+1+1=6 $

$\therefore \quad$ Required probability $=\frac{6}{52}=\frac{3}{26}$

(iv) When a pair of dice is rolled total sample parts are 36 . Out of which $(1,5),(5,1),(2,4)$, $(4,2)$ and $(3,3)$.

$\therefore \quad$ Required probability $=\frac{5}{36}$

Solution

(a) In a non-leap year’ there are 365 days which have 52 weeks and 1 day. If this day is a Tuesday or Wednesday, then the year will have 53 Tuesday or 53 Wednesday.

$\therefore \quad$ Required probability $=\frac{1}{7}$

• Option (b) $\frac{2}{7}$ is incorrect because it overestimates the probability. In a non-leap year, there is only one extra day, and the probability of this day being either a Tuesday or a Wednesday is $\frac{2}{7}$, but this does not account for the fact that we are looking for the probability of having exactly 53 Tuesdays or 53 Wednesdays, not both.

• Option (c) $\frac{3}{7}$ is incorrect because it significantly overestimates the probability. The extra day in a non-leap year can only be one of the seven days of the week, so the probability of it being either a Tuesday or a Wednesday is $\frac{2}{7}$, not $\frac{3}{7}$.

• Option (d) None of these is incorrect because the correct probability is indeed $\frac{1}{7}$, as calculated. The extra day in a non-leap year has an equal chance of being any day of the week, so the probability of it being a Tuesday or a Wednesday is $\frac{1}{7}$ for each day.

Solution

(b) Since, the set of three consecutive numbers from 1 to 20 are $123,234,345, \ldots \ldots, 18$, 19, 20 i.e., 18.

$ \begin{aligned} P(\text { numbers are consecutive }) & =\frac{18}{{ }^{20} C_3}=\frac{18}{1140}=\frac{3}{190} \\ P(\text { three numbers are not consecutive }) & =1-\frac{3}{190}=\frac{187}{190} \end{aligned} $

• Option (a) $\frac{186}{190}$ is incorrect because it does not match the correct probability calculation. The correct probability of the three numbers not being consecutive is $\frac{187}{190}$, as derived from the complement of the probability of them being consecutive.

• Option (c) $\frac{188}{190}$ is incorrect because it overestimates the probability. The correct probability, as calculated, is $\frac{187}{190}$.

• Option (d) $\frac{18}{{ }^{20} C_3}$ is incorrect because it represents the probability of the three numbers being consecutive, not the probability of them not being consecutive. The correct probability of the numbers not being consecutive is the complement of this value, which is $\frac{187}{190}$.

Solution(c) Since, in a back of 52 cards 26 are red colour and 26 are black colour.

$\therefore P$ (both cards of opposite colour) $=\frac{26}{52} \times \frac{26}{51}+\frac{26}{52} \times \frac{26}{51}$

$ =2 \times \frac{26}{52} \times \frac{26}{51}=\frac{26}{51} $

• Option (a) $\frac{29}{52}$ is incorrect because it does not correctly represent the probability calculation for the event of drawing two cards of different colors. The correct probability involves the product of fractions representing the chances of drawing one card of each color, not a simple fraction like $\frac{29}{52}$.

• Option (b) $\frac{1}{2}$ is incorrect because it oversimplifies the probability calculation. The actual probability involves specific combinations of drawing one red and one black card, which results in a more precise fraction rather than an even 50% chance.

• Option (d) $\frac{27}{51}$ is incorrect because it does not accurately reflect the probability of drawing two cards of different colors. The correct calculation involves the product of the probabilities of drawing one card of each color, which results in $\frac{26}{51}$, not $\frac{27}{51}$.

Solution

(c) If two persons sit next to each other, then consider these two person as 1 group. Now, we have to arrange 6 persons.

$ \begin{aligned} & \therefore \quad \text { Number of arrangement }=2 ! \times 6 ! \\ & \text { Total number of arrangement of } 7 \text { persons }=7 ! \end{aligned} $

$ \text { Required probability }=\frac{2 ! 6 !}{7 !}=\frac{2}{7} $

• Option (a) $\frac{1}{3}$: This option is incorrect because it does not correctly account for the number of ways to arrange the seven persons. The correct probability calculation involves considering the two particular persons as a single unit and then arranging the remaining persons, which results in a probability of $\frac{2}{7}$, not $\frac{1}{3}$.

• Option (b) $\frac{1}{6}$: This option is incorrect because it underestimates the probability. The calculation for the probability of two particular persons sitting next to each other involves treating them as a single unit and arranging the remaining persons, leading to a probability of $\frac{2}{7}$, not $\frac{1}{6}$.

• Option (d) $\frac{1}{2}$: This option is incorrect because it overestimates the probability. The correct approach involves treating the two particular persons as a single unit and arranging the remaining persons, which results in a probability of $\frac{2}{7}$, not $\frac{1}{2}$.

Solution

(d) We have, to form four-digit number using the digit $0,2,3$ and 5 which are divisible by 5 .

If 0 is fixed at units place $=3 \times 2 \times 1=6$

If 5 is fixed at units place $=2 \times 2 \times 1=4$

Total four-digit numbers divisible by $5=6+4=10$

$\therefore \quad$ Required probability $=\frac{10}{18}=\frac{5}{9}$

• Option (a) $\frac{1}{5}$: This option is incorrect because it does not accurately represent the ratio of four-digit numbers divisible by 5 to the total number of four-digit numbers that can be formed using the digits 0, 2, 3, and 5 without repetition. The correct ratio is $\frac{5}{9}$, not $\frac{1}{5}$.

• Option (b) $\frac{4}{5}$: This option is incorrect because it significantly overestimates the probability. The correct probability is $\frac{5}{9}$, which is much less than $\frac{4}{5}$.

• Option (c) $\frac{1}{30}$: This option is incorrect because it significantly underestimates the probability. The correct probability is $\frac{5}{9}$, which is much greater than $\frac{1}{30}$.

Solution

(a) For mutually exclusive events,

$ \begin{aligned} & P(A \cap B)=0 \\ & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \Rightarrow \quad P(A \cup B)=P(A)+P(B) \\ & \Rightarrow \quad P(A)+P(B) \leq 1 \\ & \Rightarrow \quad P(A)+1-P(\bar{B}) \leq 1 \quad[\because P(B)=1-P(\bar{B})] \\ & \therefore \quad P(A) \leq P(\bar{B}) \end{aligned} $

• Option (b) $P(A) \geq P(\bar{B})$: This option is incorrect because, as shown in the solution, ( P(A) \leq P(\bar{B}) ). Since ( P(A) \leq P(\bar{B}) ), it cannot be true that ( P(A) \geq P(\bar{B}) ).

• Option (c) $P(A) < P(\bar{B})$: This option is incorrect because ( P(A) \leq P(\bar{B}) ) does not necessarily imply ( P(A) < P(\bar{B}) ). It is possible for ( P(A) ) to be equal to ( P(\bar{B}) ), so the strict inequality ( P(A) < P(\bar{B}) ) is not always true.

• Option (d) None of these: This option is incorrect because option (a) is correct, as demonstrated in the solution. Therefore, it is not true that none of the options are correct.

Solution

(a) Given, $P(A \cup B)=P(A \cap B)$

$ \begin{aligned} & P(A)+P(B)-P(A \cap B)=P(A \cap B) \\ & \Rightarrow \quad[P(A)-P(A \cap B)]+[P(B)-P(A \cap B)]=0 \\ & \text { But } \quad P(A)-P(A \cap B) \geq 0 \\ & \Rightarrow \quad P(A)-P(A \cap B)=0 \\ & \text { and } \quad P(B)-P(A \cap B)=0 \end{aligned} $

$ \text { and } \quad P(B)-P(A \cap B) \geq 0 \quad[\because P(A \cap B) \leq P(A) \text { or } P(B)] $

[since, sum of two non-negative numbers can be zero only when these numbers aree zero]

$ \begin{matrix} \Rightarrow & P(A)=P(A \cap B) \\ \text { and } & P(B)=P(A \cap B) \\ \therefore & P(A)=P(B) \end{matrix} $

• Option (b) $P(A)>P(B)$: This option is incorrect because, from the given condition (P(A \cup B) = P(A \cap B)), we derived that (P(A) = P(B)). Therefore, it is not possible for (P(A)) to be greater than (P(B)).

• Option (c) $P(A)<P(B)$: This option is incorrect because, similar to the reasoning for option (b), we derived that (P(A) = P(B)) from the given condition (P(A \cup B) = P(A \cap B)). Therefore, it is not possible for (P(A)) to be less than (P(B)).

• Option (d) None of these: This option is incorrect because we have shown that (P(A) = P(B)) is a valid conclusion from the given condition (P(A \cup B) = P(A \cap B)). Therefore, option (a) is correct, and “None of these” cannot be the correct answer.

Solution

(c) If all the girls sit together, then considered it as 1 group.

$\therefore$ Arrangement of $6+1=7$ person in a row is 7 ! and the girls interchanges their seets in 6 ! ways.

$\therefore \quad$ Required probability $=\frac{6 ! 7 !}{12 !}=\frac{1}{132}$

• Option (a) $\frac{1}{432}$ is incorrect because it does not account for the correct number of arrangements of the boys and girls. The correct calculation involves the factorial of 7 and 6, not 432.

• Option (b) $\frac{12}{431}$ is incorrect because it is not a simplified fraction and does not represent the correct probability. The correct probability should be a fraction derived from factorial calculations, not a ratio like 12/431.

• Option (d) None of these is incorrect because there is a correct option provided, which is (c) $\frac{1}{132}$.

Solution

(b) Total number of alphabet in the word probability $=11$

$ \begin{aligned} \text { Number of vowels } & =4 \\ P(\text { letter is vowel }) & =\frac{4}{11} \end{aligned} $

• Option (a) $\frac{1}{3}$: This option is incorrect because the probability of selecting a vowel from the word ‘PROBABILITY’ is not $\frac{1}{3}$. The correct probability is $\frac{4}{11}$, as there are 4 vowels out of 11 letters. $\frac{1}{3}$ would imply that there are 3 or 9 vowels out of 9 or 27 letters, which is not the case here.

• Option (c) $\frac{2}{11}$: This option is incorrect because it underestimates the number of vowels in the word ‘PROBABILITY’. There are 4 vowels (A, O, I, I) in the word, not 2. Therefore, the probability should be $\frac{4}{11}$, not $\frac{2}{11}$.

• Option (d) $\frac{3}{11}$: This option is incorrect because it also underestimates the number of vowels in the word ‘PROBABILITY’. There are 4 vowels in the word, not 3. Hence, the correct probability is $\frac{4}{11}$, not $\frac{3}{11}$.

Solution

(c) Given,

$ P(A \text { fail })=0.2 $

and

$P(B$ fail $)=0.3$

$\therefore$

$ \begin{aligned} P(\text { either } A \text { or } B \text { fail }) & \leq P(A \text { fail })+P(B \text { fail }) \\ & \leq 0.2+0.3 \\ & \leq 0.5 \end{aligned} $

• Option (a) >0.5: This option is incorrect because the maximum possible probability that either A or B fails is the sum of their individual probabilities, which is 0.2 + 0.3 = 0.5. Therefore, the probability cannot be greater than 0.5.

• Option (b) 0.5: This option is incorrect because the probability that either A or B fails could be less than 0.5 if there is some overlap (i.e., if there is a chance that both A and B fail simultaneously). The given information does not specify that A and B failing are mutually exclusive events, so the probability could be less than 0.5.

• Option (d) 0: This option is incorrect because there is a non-zero probability that either A or B fails. Given that P(A fails) = 0.2 and P(B fails) = 0.3, there is a definite chance that at least one of them will fail, making the probability greater than 0.

Solution

(c) Given,

$ \begin{aligned} P(A \cup B) & =0.6 \text { and } P(A \cap B)=0.2 \\ P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ 0.6 & =P(A)+P(B)-0.2 \\ (A)+P(B) & =0.8 \\ (\bar{A})+P(\bar{B}) & =1-P(A)+1-P(B) \\ & =2-[P(A)+P(B)] \\ & =2-0.8=1.2 \end{aligned} $

$ \begin{matrix} \Rightarrow & 0.6 =P(A)+P(B)-0.2 \\ \Rightarrow & P(A)+P(B) =0.8 \\ \therefore & P(\bar{A})+P(\bar{B}) =1-P(A)+1-P(B) \end{matrix} $

• Option (a) 0.4: This option is incorrect because it does not account for the correct calculation of the probabilities of the complements of events (A) and (B). The correct calculation shows that (P(\bar{A}) + P(\bar{B}) = 1.2), not 0.4.

• Option (b) 0.8: This option is incorrect because it represents the sum of the probabilities of events (A) and (B) occurring, i.e., (P(A) + P(B) = 0.8). However, the question asks for the sum of the probabilities of the complements of (A) and (B), which is different and equals 1.2.

• Option (d) 1.6: This option is incorrect because it overestimates the sum of the probabilities of the complements of events (A) and (B). The correct calculation shows that (P(\bar{A}) + P(\bar{B}) = 1.2), not 1.6.

Solution

(b) If $M$ and $N$ are any two events.

$\therefore \quad P(M \cup N)=P(M)+P(N)-P(M \cap N)$

• Option (a): This option is incorrect because it subtracts twice the probability of the intersection of events (M) and (N). The correct formula only requires subtracting the intersection once to avoid double-counting the overlap.

• Option (c): This option is incorrect because it adds the probability of the intersection of events (M) and (N) instead of subtracting it. Adding the intersection would overestimate the probability of the union of the two events.

• Option (d): This option is incorrect because it adds twice the probability of the intersection of events (M) and (N). The correct formula should subtract the intersection once, not add it, to avoid overestimating the probability of the union.

Solution

False

$P$ (to see girafee) $=0.72$

$P$ (to see bear) $=0.84$

$P$ (to see giraffee and bear) $=0.52$

$P($ to see giraffee or bear $)=P($ giraffee $)+P($ bear $)-P($ giraffee and bear $)$

$=0.72+0.84-0.52$

$=1.04$

which is not possible. Hence statement is false.

• The probability that a person visiting a zoo will see the giraffe is 0.72, the probability that he will see the bears is 0.84, and the probability that he will see both is 0.52.

The reason why the statement is false is that the calculated probability of seeing either the giraffe or the bear exceeds 1, which is not possible in probability theory. Specifically:

$P($ to see giraffe or bear $) = P($ giraffe $) + P($ bear $) - P($ giraffe and bear $)$

$= 0.72 + 0.84 - 0.52$

$= 1.04$

Since probabilities cannot exceed 1, the given probabilities are inconsistent, making the statement false.

Solution

False

Let $\quad A=$ Student will pass examination

$B=$ Student will getting compartment

$P(A)=0.73$ and $P(A$ or $B)=0.96$ and $P(B)=0.13$

$\therefore \quad P(A$ or $B)=P(A)+P(B)=0.73+0.13=0.86$

But $\quad P(A$ or $B)=0.96$

Hence, it is false statement.

• The given probabilities do not add up correctly according to the rules of probability. Specifically, the probability of the student either passing or getting a compartment should be the sum of the individual probabilities minus the probability of both events happening simultaneously. The calculation provided in the answer does not account for the overlap between the two events, which is why the statement is false.

Solution

False

Sum of these probabilities must be equal to 1 .

$ \begin{aligned} P(0)+P(1) & +P(2)+P(3)+P(4)+P(5) \\ & =0.12+0.25+0.36+0.14+0.08+0.11=1.06 \end{aligned} $

which is greater than 1 ,

So, it is false statement.

• The sum of the probabilities must equal 1 for a valid probability distribution. In this case, the sum is 1.06, which is greater than 1, indicating an error in the given probabilities.

Solution

False

Here, $\quad P(A)=0.5, P(A \cap B) \leq 0.3$

Now, $\quad P(A) \times P(B) \leq 0.3$

$\Rightarrow \quad 0.5 \times P(B) \leq 0.3$

$\begin{matrix} \Rightarrow & P(B) \leq 0.6\end{matrix} $

Hence, it is false statement.

• The probability that both $A$ and $B$ are selected is at most 0.3, which means $P(A \cap B) \leq 0.3$. Given that $P(A) = 0.5$, we can use the formula for the intersection of two events: $P(A \cap B) = P(A) \times P(B)$ if $A$ and $B$ are independent. However, even if they are not independent, $P(A \cap B) \leq P(A) \times P(B)$.

• To check if $P(B) = 0.7$ is possible, we use the inequality $P(A \cap B) \leq 0.3$ and $P(A) = 0.5$:

  [ P(A) \times P(B) \leq 0.3 ]

  Substituting $P(A) = 0.5$:

  [ 0.5 \times P(B) \leq 0.3 ]

  Solving for $P(B)$:

  [ P(B) \leq \frac{0.3}{0.5} = 0.6 ]

• Therefore, $P(B) = 0.7$ is not possible because it exceeds the maximum allowable value of 0.6. Hence, the statement that the probability of $B$ getting selected is 0.7 is false.

Solution

True

$ P(A \cap B) \leq P(A) $

Hence, it is true statement.

• The statement “The probability of intersection of two events ( A ) and ( B ) is always greater than or equal to those favourable to the event ( A )” is incorrect because the probability of the intersection of two events ( P(A \cap B) ) can never be greater than the probability of either event occurring individually. This is due to the fact that ( P(A \cap B) ) represents the probability that both events occur simultaneously, which is inherently limited by the probability of the less likely event.

• The statement “The probability of intersection of two events ( A ) and ( B ) is always equal to those favourable to the event ( A )” is incorrect because ( P(A \cap B) ) is only equal to ( P(A) ) if event ( B ) is a certain event (i.e., ( P(B) = 1 )) and ( A ) is a subset of ( B ). In general, ( P(A \cap B) ) is less than or equal to ( P(A) ).

• The statement “The probability of intersection of two events ( A ) and ( B ) is always less than or equal to those favourable to the event ( B )” is incorrect because while it is true that ( P(A \cap B) \leq P(B) ), this does not address the comparison with ( P(A) ). The original question specifically asks about the relationship between ( P(A \cap B) ) and ( P(A) ), not ( P(B) ).

• The statement “The probability of intersection of two events ( A ) and ( B ) is always greater than or equal to those favourable to the event ( B )” is incorrect because ( P(A \cap B) ) can never be greater than ( P(B) ). The intersection probability ( P(A \cap B) ) is limited by the probabilities of both events occurring, and thus it cannot exceed the probability of either individual event.

Solution

False

Here,

and

$ \begin{aligned} P(A) & =0.7 \\ P(B) & =0.3 \\ P(A \cap B) & =P(A) \times P(B) \\ & =0.7 \times 0.3=0.21 \end{aligned} $

Hence, it is false statement.

• The given probability of the occurrence of both events $A$ and $B$ is 0.4, but the calculated probability assuming independence is 0.21. This discrepancy indicates that the events $A$ and $B$ are not independent, which makes the statement false.
• The calculation $P(A \cap B) = P(A) \times P(B) = 0.7 \times 0.3 = 0.21$ is only valid if events $A$ and $B$ are independent. Since the given $P(A \cap B) = 0.4$ does not match this, the assumption of independence is incorrect.
• The statement “The probability of occurrence of both is 0.4” contradicts the calculated value under the assumption of independence, reinforcing that the events are not independent and making the statement false.

Solution

True

Since, these two events not related to the same sample space.

So, sum of probabilities of two students getting distinction in their final examination may be 1.2 .

Hence, it is true statement.

• The sum of probabilities of two students getting distinction in their final examinations cannot exceed 1 if they are from the same sample space, as the total probability in a single sample space must always equal 1. Therefore, if the events are related to the same sample space, the sum of their probabilities cannot be 1.2.

• If the events are mutually exclusive (i.e., they cannot both happen at the same time), the sum of their probabilities should still not exceed 1. A sum of 1.2 would imply that the events are not mutually exclusive and are not from the same sample space.

• The sum of probabilities of independent events (events that do not affect each other) should also not exceed 1 if they are from the same sample space. If the sum is 1.2, it indicates that the events are not independent or not from the same sample space.

• The sum of probabilities of two events in a single sample space must always be less than or equal to 1. A sum of 1.2 suggests that the events are not from the same sample space or there is an error in the probability calculation.

Solution

$\quad P$ (lossing) $=1-(0.77+0.08)=0.15$

Solution

$ P(e_1)+P(e_2)+P(e_3)+P(e_4)=1 $

$ \begin{matrix} \Rightarrow & 0.1+0.5+0.1+P(e_4) =1 \\ \Rightarrow & 0.7+P(e_4) =1 \\ \therefore & P(e_4) =0.3 \end{matrix} $

Solution

Here,

and

$ \begin{aligned} & S=\lbrace1,2,3,4,5,6\rbrace \\ & E=\lbrace1,3,5\rbrace \\ & \bar{E}=S-E=\lbrace2,4,6\rbrace \end{aligned} $

$ \therefore $

Solution

$P(A \cap \bar{B})=P(A)-P(A \cap B)=0.3-0.1=0.2$

Solution

$\quad P(\bar{A} \cap \bar{B})=P(\overline{A \cup B})=1-P(A \cup B)$

$ \begin{aligned} & =1-[P(A)+P(B)] \quad[\text { since, } A \text { and } B \text { are mutually exclusive }] \\ & =1-(0.5+0.3)=1-0.8=0.2 \end{aligned} $

Solution

(i) 0.95 is very likely to happen, so it is close to 1 .

(ii) 0.02 very little chance of happening because probability is very low.

(iii) -0.3 an incorrect assignment because probability of any events lie between 0 and 1 .

(iv) 0.5 , as much chance of happening as not because sum of chances of happening and not happening is zero.

(v) 0 , no chance of happening.

Solution

(i) If $E_1$ and $E_2$ are two mutually exclusive event, then $E_1 \cap E_2=\varphi$.

(ii) If $E_1$ and $E_2$ are mutually exclusive and exhaustive events, then $E_1 \cap E_2=\varphi$ and $E_1 \cup E_2=S$.

(iii) If $E_1$ and $E_2$ have common outcomes, then $(E_1-E_2) \cup(E_1 \cap E_2)=E_1$

(iv) If $E_1$ and $E_2$ are two events such that $E_1 \subset E_2 \Rightarrow E_1 \cap E_2=E_1$