Answer :

(a)

Here, each of the numbers $p(\omega _{i})$ is positive and less than 1.

Sum of probabilities

$=p(\omega_1)+p(\omega_2)+p(\omega_3)+p(\omega_4)+p(\omega_5)+p(\omega_6)+p(\omega_7)$

$=0.1+0.01+0.05+0.03+0.01+0.2+0.6$

$=1$

Thus, the assignment is valid.

(b)

Here, each of the numbers $p(\omega _{i})$ is positive and less than 1.

Sum of probabilities

$=p(\omega_1)+p(\omega_2)+p(\omega_3)+p(\omega_4)+p(\omega_5)+p(\omega_6)+p(\omega_7)$

$=\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}=7 \times \dfrac{1}{7}=1$

Thus, the assignment is valid.

(c)

Here, each of the numbers is positive and less than 1.

Sum of probabilities

$=p(\omega_1)+p(\omega_2)+p(\omega_3)+p(\omega_4)+p(\omega_5)+p(\omega_6)+p(\omega_7)$

$=0.1+0.2+0.3+0.4+0.5+0.6+0.7$

$=2.8 \neq 1$

Thus, the assignment is not valid.

(d)

Here, $p(\omega _1)$ and $p( \omega _5)$ are negative.

Hence, the assignment is not valid.

(e)

Here,

$ p(\omega_7)=\dfrac{15}{14}>1 $

Hence, the assignment is not valid.

Answer :

When a coin is tossed twice, the sample space is given by S={H H, H T, T H, T T}

Let $A$ be the event of the occurrence of at least one tail.

Accordingly, A={HT, TH, TT}

$\therefore P(A)=\dfrac{\text{ Number of outcomes favourable to } A}{\text{ Total number of possible outcomes }}$

$ \begin{aligned} & =\dfrac{n(A)}{n(S)} \\ & =\dfrac{3}{4} \end{aligned} $

Answer :

The sample space of the given experiment is given by

S={1,2,3,4,5,6}

(i) Let $A$ be the event of the occurrence of a prime number.

Accordingly, A={2,3,5}

$\therefore P(A)=\dfrac{\text{ Number of outcomes favourable to A }}{\text{ Total number of possible outcomes }}=\dfrac{n(A)}{n(S)}=\dfrac{3}{6}=\dfrac{1}{2}$

(ii) Let $B$ be the event of the occurrence of a number greater than or equal to 3 . Accordingly, B={3,4,5,6}

$\therefore P(B)=\dfrac{\text{ Number of outcomes favourable to B }}{\text{ Total number of possible outcomes }}=\dfrac{n(B)}{n(S)}=\dfrac{4}{6}=\dfrac{2}{3}$

(iii) Let $C$ be the event of the occurrence of a number less than or equal to one. Accordingly, C={1}

$\therefore P(C)=\dfrac{\text{ Number of outcomes favourable to } C}{\text{ Total number of possible outcomes }}=\dfrac{n(C)}{n(S)}=\dfrac{1}{6}$

(iv) Let $D$ be the event of the occurrence of a number greater than 6 .

Accordingly, D = no elements

$\therefore P(D)=\dfrac{\text{ Number of outcomes favourable to } D}{\text{ Total number of possible outcomes }}=\dfrac{n(D)}{n(S)}=\dfrac{0}{6}=0$

(v) Let $E$ be the event of the occurrence of a number less than 6 .

Accordingly, E={1,2,3,4,5}

$\therefore P(E)=\dfrac{\text{ Number of outcomes favourable to } E}{\text{ Total number of possible outcomes }}=\dfrac{n(E)}{n(S)}=\dfrac{5}{6}$

Answer :

(a) When a card is selected from a pack of 52 cards, the number of possible outcomes is 52 i.e., the sample space contains 52 elements.

Therefore, there are 52 points in the sample space.

(b) Let $A$ be the event in which the card drawn is an ace of spades.

Accordingly, $n(A)=1$

$\therefore P(A)=\dfrac{\text{ Number of outcomes favourable to } A}{\text{ Total number of possible outcomes }}=\dfrac{n(A)}{n(S)}=\dfrac{1}{52}$

(c) (i)Let $E$ be the event in which the card drawn is an ace.

Since there are 4 aces in a pack of 52 cards, $n(E)=4$

$\therefore P(E)=\dfrac{\text{ Number of outcomes favourable to } E}{\text{ Total number of possible outcomes }}=\dfrac{n(E)}{n(S)}=\dfrac{4}{52}=\dfrac{1}{13}$

(ii)Let $F$ be the event in which the card drawn is black.

Since there are 26 black cards in a pack of 52 cards, $n(F)=26$

$\therefore P(F)=\dfrac{\text{ Number of outcomes favourable to } F}{\text{ Total number of possible outcomes }}=\dfrac{n(F)}{n(S)}=\dfrac{26}{52}=\dfrac{1}{2}$

Answer :

Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1, 2, 3, 4,5 , and 6 , the sample space is given by

S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Accordingly, $n(S)=12$

(i) Let $A$ be the event in which the sum of numbers that turn up is 3 .

Accordingly, A={(1,2)}

$\therefore P(A)=\dfrac{\text{ Number of outcomes favourable to A }}{\text{ Total number of possible outcomes }}=\dfrac{n(A)}{n(S)}=\dfrac{1}{12}$

(ii) Let $B$ be the event in which the sum of numbers that turn up is 12 .

Accordingly, B={(6,6)}

$\therefore P(B)=\dfrac{\text{ Number of outcomes favourable to B }}{\text{ Total number of possible outcomes }}=\dfrac{n(B)}{n(S)}=\dfrac{1}{12}$

Answer :

There are four men and six women on the city council.

As one council member is to be selected for a committee at random, the sample space contains $10(4+6)$ elements.

Let $A$ be the event in which the selected council member is a woman.

Accordingly, $n(A)=6$

$\therefore P(A)=\dfrac{\text{ Number of outcomes favourable to } A}{\text{ Total number of possible outcomes }}=\dfrac{n(A)}{n(S)}=\dfrac{6}{10}=\dfrac{3}{5}$

Answer :

Since the coin is tossed four times, there can be a maximum of 4 heads or tails.

When 4 heads turns up, $Rs 1+Rs 1+Rs 1+Rs 1=Rs 4$ is the gain.

When 3 heads and 1 tail turn up, $Rs 1+Rs 1+Rs 1$- Rs $1.50=Rs 3$ - Rs $1.50=Rs 1.50$ is the gain.

When 2 heads and 2 tails turns up, $Rs 1+Rs 1$ - Rs 1.50 - Rs $1.50=$ - Rs 1, i.e., $Rs 1$ is the loss.

When 1 head and 3 tails turn up, Rs 1 - Rs 1.50 - Rs 1.50 - Rs $1.50=$- Rs 3.50, i.e., Rs 3.50 is the loss.

When 4 tails turn up, - Rs 1.50 - Rs 1.50 - Rs 1.50 - Rs 1.50 = - Rs 6.00, i.e., Rs 6.00 is the loss.

There are $2^{4}=16$ elements in the sample space $S$, which is given by:

S={ HHHH, HHHT, HHTH, HTHH, THHH}

{HHTT, HTTH, TTHH, HTHT, THTH, THHT}

{HTTT, THTT, TTHT, TTTH}

{TTTT }

$\therefore n(S)=16$

The person wins Rs 4.00 when 4 heads turn up, i.e., when the event $\{HHHH\}$ occurs.

$\therefore$ Probability (of winning Rs 4.00) $=\dfrac{1}{16}$

The person wins Rs 1.50 when 3 heads and one tail turn up, i.e., when the event $\{HHHT, HHTH, HTHH, THHH\}$ occurs.

$\therefore$ Probability (of winning Rs 1.50) $=\dfrac{4}{16}=\dfrac{1}{4}$

The person loses Rs 1.00 when 2 heads and 2 tails turn up, i.e., when the event $\{ HHTT, HTTH, TTHH, HTHT, THTH, THHT \}$ occurs.

$\therefore$ Probability (of losing Rs 1.00) $=\dfrac{6}{16}=\dfrac{3}{8}$

The person loses Rs 3.50 when 1 head and 3 tails turn up, i.e., when the event $\{ HTTT, THTT, TTHT, TTTH\}$ occurs.

Probability (of losing Rs 3.50) $=\dfrac{4}{16}=\dfrac{1}{4}$

The person loses Rs 6.00 when 4 tails turn up, i.e., when the event $\{ TTTT\}$ occurs.

Probability (of losing Rs 6.00) $=\dfrac{1}{16}$

Answer :

When three coins are tossed once, the sample space is given by

$S=\{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}$

$\therefore$ Accordingly, $n(S)=8$

It is known that the probability of an event $A$ is given by

$P(A)=\dfrac{\text{ Number of outcomes favourable to } A}{\text{ Total number of possible outcomes }}=\dfrac{n(A)}{n(S)}$

(i) Let $B$ be the event of the occurrence of 3 heads. Accordingly, $B=\{HHH\}$

$\therefore P(B)=\dfrac{n(B)}{n(S)}=\dfrac{1}{8}$

(ii) Let $C$ be the event of the occurrence of 2 heads. Accordingly, C={HHT, HTH, THH}

$\therefore P(C)=\dfrac{n(C)}{n(S)}=\dfrac{3}{8}$

(iii) Let $D$ be the event of the occurrence of at least 2 heads.

Accordingly, D={HHH, HHT, HTH, THH}

$\therefore P(D)=\dfrac{n(D)}{n(S)}=\dfrac{4}{8}=\dfrac{1}{2}$

(iv) Let $E$ be the event of the occurrence of at most 2 heads.

Accordingly, E={HHT, HTH, THH, HTT, THT, TTH, TTT}

$\therefore P(E)=\dfrac{n(E)}{n(S)}=\dfrac{7}{8}$

(v) Let $F$ be the event of the occurrence of no head.

Accordingly, F={TTT}

$\therefore P(F)=\dfrac{n(F)}{n(S)}=\dfrac{1}{8}$

(vi) Let $G$ be the event of the occurrence of 3 tails.

Accordingly, G={TTT}

$\therefore P(G)=\dfrac{n(G)}{n(S)}=\dfrac{1}{8}$

(vii) Let $H$ be the event of the occurrence of exactly 2 tails.

Accordingly, H={HTT, THT, TTH}

$\therefore P(H)=\dfrac{n(H)}{n(S)}=\dfrac{3}{8}$

(viii) Let I be the event of the occurrence of no tail.

Accordingly, I={HHH}

$\therefore P(I)=\dfrac{n(I)}{n(S)}=\dfrac{1}{8}$

(ix) Let $J$ be the event of the occurrence of at most 2 tails.

Accordingly, I = { HHH, HHT, HTH, THH, HTT, THT, TTH}

$\therefore P(J)=\dfrac{n(J)}{n(S)}=\dfrac{7}{8}$

Answer :

It is given that $P(A)=\dfrac{2}{11}$.

Accordingly, $P(not A)=1$- $P(A)=1-\dfrac{2}{11}=\dfrac{9}{11}$

Answer :

There aRs 13 letters in the word ASSASSINATION.

$\therefore$ Hence, $n(S)=13$

(i) There are 6 vowels in the given word.

$\therefore$ Probability (vowel) $=\dfrac{6}{13}$ (ii) There are 7 consonants in the given word.

$\therefore$ Probability (consonant) $=\dfrac{7}{13}$

Answer :

Total number of ways in which one can choose six different numbers from 1 to 20

$ { }^{20}C_6 = \dfrac{\lfloor{20}}{{\lfloor{6}} \lfloor{20-7} } = \dfrac{\lfloor{20}}{{\lfloor{6}} \lfloor{14} }$

$=\dfrac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}=38760$

Hence, there are 38760 combinations of 6 numbers.

Out of these combinations, one combination is already fixed by the lottery committee.

$\therefore$ Required probability of winning the prize in the game $=\dfrac{1}{38760}$

Answer :

(i) $ P(A)=0.5, P(B)=0.7, P(A \cap B)=0.6$

It is known that if $E$ and $F$ are two events such that $E$ âŠ, $F$, then $P(E) \hat{a} \% \propto P(F)$.

However, here, $P(A \cap B)>P(A)$.

Hence, $P(A)$ and $P(B)$ are not consistently defined.

(ii) $P(A)=0.5, P(B)=0.4$

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

$ 0.8 = 0.5 +0.4 - P(A \cap B) $

$ P(A \cap B) = 0.5 + 0.4 - 0.8 $

$ = 0.1 $

Since $ P(A \cap B) < P(A) \& P(B), $

the given data is consistent

Answer :

(i) Here,

$ P(A)=\dfrac{1}{3}, P(B)=\dfrac{1}{5}, P(A \cap B)=\dfrac{1}{15} $

We know that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\therefore P(A \cup B)=\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{15}=\dfrac{5+3-1}{15}=\dfrac{7}{15}$

(ii) Here, $P(A)=0.35, P(A \cap B)=0.25, P(A \cup B)=0.6$

We know that $P(A \cup B)=P(A)+P(B) - P(A \cap B)$

$\therefore 0.6=0.35+P(B) - 0.25$

$\Rightarrow P(B)=0.6$ - $0.35+0.25$

$\Rightarrow P(B)=0.5$

(iii)Here, $P(A)=0.5, P(B)=0.35, P(A \cap B)=0.7$

We know that $P(A \cup B)=P(A)+P(B) - P(A \cap B)$

$\therefore 0.7=0.5+0.35$ - $P(A \cap B)$

$\Rightarrow P(A \cap B)=0.5+0.35$ - 0.7

$\Rightarrow P(A \cap B)=0.15$

Answer :

Here, $P(A)=\dfrac{3}{5}{}, P(B)=\dfrac{1}{5}$

For mutually exclusive events $A$ and $B$,

$P(A$ or $B)=P(A)+P(B)$

$\therefore P(A$ or $B)=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}$

$ =\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5} $

Answer :

Here, $P(E)=\dfrac{1}{4}, P(F)=\dfrac{1}{2}$, and $P(E$ and $F)=\dfrac{1}{8}$

(i) We know that $P(E \text{ or } F)=P(E)+P(F) - P(E and F)$

$\therefore P(E$ or $F)=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{2+4-1}{8}=\dfrac{5}{8}$

(ii) From (i), $P(E$ or $F)=P(E \cup F)=\dfrac{5}{8}$

We have $(E \cup F)^{\prime}=(E^{\prime} \cap F^{\prime}) \quad[$ By De Morgan’s law]

$\therefore P(E^{\prime} \cap F^{\prime})=P(E \cup F)^{\prime}$

Now, $P(E \cup F)^{\prime}=1-P(E \cup F)=1-\dfrac{5}{8}=\dfrac{3}{8}$

$\therefore P(E^{\prime} \cap F^{\prime})=\dfrac{3}{8}$

Thus, $P(not E$ and $not F)=\dfrac{3}{8}$

Answer :

It is given that $P$ (not $E$ or not $F)=0.25$ i.e., $P(E^{\prime} \cup F^{\prime})=0.25$

$\Rightarrow P(E \cap F)^{\prime}=0.25$

$ [E^{\prime} \cup F^{\prime}=(E \cap F)^{\prime}] $

Now, $P(E \cap F)=1-P(E \cap F)^{\prime}$

$\Rightarrow P(E \cap F)=1-0.25$

$\Rightarrow P(E \cap F)=0.75 \neq 0$

$\Rightarrow E \cap F \neq \phi$

Thus, $E$ and $F$ are not mutually exclusive.

Answer :

It is given that $P(A)=0.42, P(B)=0.48, P(A$ and $B)=0.16$

(i) $P(not A)=1-P(A)=1-0.42=0.58$

(ii) $P($ not $B)=1-P(B)=1-0.48=0.52$

(iii) We know that $P(A$ or $B)=P(A)+P(B)-P(A$ and $B)$

$ P(A$ or $B)=0.42+0.48-0.16=0.74$

Answer :

Let $A$ be the event in which the selected student studies Mathematics and $B$ be the event in which the selected student studies Biology.

Accordingly, $P(A)=40 \%=\dfrac{40}{100}=\dfrac{2}{5}$

$P(B)=30 \%=\dfrac{30}{100}=\dfrac{3}{10}$

$P(A$ and $B)=10 \%=\dfrac{10}{100}=\dfrac{1}{10}$

We know that $P(A$ or $B)=P(A)+P(B)$ $P(A$ and $B)$

$\therefore P(A$ or $B)=\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{10}=\dfrac{6}{10}=0.6$

Thus, the probability that the selected student will be studying Mathematics or Biology is 0.6 .

Answer :

Let $A$ and $B$ be the events of passing first and second examinations respectively.

Accordingly, $P(A)=0.8, P(B)=0.7$ and $P(A$ or $B)=0.95$

We know that $P(A$ or $B)=P(A)+P(B)-P(A$ and $B)$

$ 0.95=0.8+0.7-P(A$ and $B)$

$P(A$ and $B)=0.8+0.7-0.95=0.55$

Thus, the probability of passing both the examinations is 0.55 .

Answer :

Let $A$ and $B$ be the events of passing English and Hindi examinations respectively.

Accordingly, $P(A$ and $B)=0.5$, $P($ not $A$ and not $B)=0.1$, i.e., $P(A^{\prime} \cap B^{\prime})=0.1$

$P(A)=0.75$

Now, $(A \cup B)^{\prime}=(A^{\prime} \cap B^{\prime}) \quad[De$ Morgan’s law]

$\therefore P(A \cup B)^{\prime}=P(A^{\prime} \cap B^{\prime})=0.1$

$P(A \cup B)=1-P(A \cup B)^{\prime}=1-0.1=0.9$

We know that $P(A$ or $B)=P(A)+P(B)$ - $P(A$ and $B)$

$ \begin{aligned} & \therefore 0.9=0.75+P(B)- 0.5 \\ & \Rightarrow P(B)=0.9 \text{ - } 0.75+0.5 \end{aligned} $

$\Rightarrow P(B)=0.65$

Thus, the probability of passing the Hindi examination is 0.65 .

Answer :

Let $A$ be the event in which the selected student has opted for NCC and B be the event in which the selected student has opted for NSS.

Total number of students $=60$

Number of students who have opted for NCC $=30$

$\therefore P(A)=\dfrac{30}{60}=\dfrac{1}{2}$

Number of students who have opted for NSS $=32$

$\therefore P(B)=\dfrac{32}{60}=\dfrac{8}{15}$

Number of students who have opted for both NCC and NSS $=24$

$\therefore P(A$ and $B)=\dfrac{24}{60}=\dfrac{2}{5}$

(i) We know that $P(A$ or $B)=P(A)+P(B)$- $P(A$ and $B)$

$\therefore P(A$ or $B)=\dfrac{1}{2}+\dfrac{8}{15}-\dfrac{2}{5}=\dfrac{15+16-12}{30}=\dfrac{19}{30}$

Thus, the probability that the selected student has opted for NCC or NSS is $\dfrac{19}{30}$.

(ii) $P($ not $A$ and not $B)$

$=P(A^{\prime}.$ and $.B^{\prime})$

$=P(A^{\prime} \cap B^{\prime})$

$=P(A \cup B)^{\prime} \quad[(A^{\prime} \cap B^{\prime})=(A \cup B)^{\prime}(.$ by De Morgan’s law $.)]$

$=1-P(A \cup B)$

$=1-P(A$ or $B)$

$=1-\dfrac{19}{30}$

$=\dfrac{11}{30}$

Thus, the probability that the selected students has neither opted for NCC nor NSS is $\dfrac{11}{30}$.

(iii) The given information can be represented by a Venn diagram as

It is clear that

Number of students who have opted for NSS but not NCC

$=n(B - A)=n(B) - n(A \cap C)=32 - 24=8$

Thus, the probability that the selected student has opted for NSS but not for NCC $=\dfrac{8}{60}=\dfrac{2}{15}$