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 $=\frac{1}{7}$

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} $

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} $

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} $

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 $=\frac{10}{18}=\frac{5}{9}$

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} $

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} $

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 $=\frac{6 ! 7 !}{12 !}=\frac{1}{132}$

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} $

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} $

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} $

Solution

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

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

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.

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.

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.

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.

Solution

True

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

Hence, it is true statement.

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.

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.

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 $E_1 \subset E_2 \Rightarrow E_1 \cap E_2=E_1$