The probability of occurrence of an event $E$ in one trial is $0.4$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2,...

The probability of safe arrival of the ship is $1 / 5$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots \ldots...

The probability that the man hits the target is $3 / 4$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots \ldots...

Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p)$, here $n=7$ As we know that the favourable outcomes of getting at...

Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n \text { and } q=(1-p)$ As we know that the favourable outcomes of getting at least...

Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p)$ As the die is thrown 5 times the total number of outcomes will be...

Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . \mathrm{n}$ and $\mathrm{q}=(1-\mathrm{p})$ As the coin is tossed 8 times the total...

Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots \ldots . .$ and $q=(1-p)$ As the coin is tossed 5 times the total number of outcomes will...

Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p)$ As the coin is tossed 5 times the total number of outcomes will be...

Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots \ldots$ and $q=(1-p)$ As the coin is thrown 6 times the total number of outcomes will be...

A die is tossed twice, The probability of getting a 4,5 or 6 in the first trial is $3 / 6=\mathrm{P}(\mathrm{A})$ The probability of getting a $1,2,3$ or 4 in the second trial is $4 / 6=P(B)$ As the...

The couple has two children and one is known to be boy, The probability that the other is boy will be $=$ $\frac{Favourable-outcome}{Total-outcome}$ Total outcomes are 3 , The first child is a boy,...

Given: $60 \%$ of the students read mathematics, $25 \%$ biology and $15 \%$ both mathematics and biology That means, Let the event A implies students reading mathematics, Let the event B implies...

The sum will be even when; both numbers are either even or odd, i.e. for both numbers to be even, the total cases ${ }^{5} \mathrm{C}_{1} \mathrm{X}^{4} \mathrm{C}_{1}$ (Both the numbers are odd)...

The die is thrown twice, So the favourable outcomes that the sum appears to be 7 are $(1,6),(2,5),(3,4),(4,3),(5,2)$ and $(6,1)$ Out of these 2 appears twice, So the probability that 2 appears at...

Given, $\begin{array}{l} P(A \cup B)=\left(\frac{5}{6}\right), P(A \cap B)=\left(\frac{1}{3}\right) \text { and } \\ P(\bar{B})=\left(\frac{1}{2}\right), P(B)=1-P(\bar{B})=1-\frac{1}{2}=\frac{1}{2}...

$\mathrm{P}(\overline{\mathrm{A}} / \overline{\mathrm{B}})=\frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})}=\frac{P(\bar{A}) P(\overline{\bar{B})}}{1-P(B)}=1-P(A)$ Hence, the correct option is a.

$\begin{array}{l} P(A)=0.4, P(B)=0.8 \text { and } \\ P(B / A)=0.6 \\ P(B / A)=\frac{P(A \cap B)}{P(A)}=0.6 \\ P(A \cap B)=0.24 \\ \Rightarrow P(A / B)=\frac{P(A \cap B)}{P(B)}=0.3 \end{array}$...

$\begin{array}{l} P(A)=\frac{1}{4}, P(B)=\frac{1}{3} \text { and } P(A \cap B)=\frac{1}{5} \\ P(\overline{(B} / \bar{A})=\frac{P(\bar{A} \cap \bar{B})}{P(\bar{A})}=\frac{1-P(A \cup...

$\begin{array}{l} P(A)=0.3, P(B)=0.2 \text { and } P(A \cap B)=0.1 \\ P(\bar{A} \cap B)=P(B)-P(A \cap B)=0.2-0.1=0.1 \end{array}$ Hemce, correct option is b.

The probability that the outcome which is either, 1,3 or 5 is prime is $=\frac{\text { Favorable outcomes }}{\text { Total outcomes }}$ Favourable outcomes $=3$ or 5 Total outcomes $=1,3$, and 5...

The probability of failure of the first component $=0.2=\mathrm{P}(\mathrm{A})$ The probability of failure of second component $=0.3=\mathrm{P}(\mathrm{B})$ The probability of failure of third...

$\begin{array}{l} P(A)=\frac{4}{5} P(B)=\frac{3}{4} P(C)=\frac{2}{3} \\ P\left(B \cap C \cap A^{\prime}\right)=P(B \cap C)-P(B \cap C \cap A) \end{array}$ As the events are independent, So, $P(B...

The probability that the problem is solved $=P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(C \cap A)+3 P(A \cap B \cap$ C) Considering independent events, $P(A \cap B)=P(A) \cdot P(B)$...

$\mathrm{P}(\mathrm{A})=$ probability that A can solve the problem $=3 / 5$ And $P(B)=$ probability that $B$ can solve the problem $=2 / 3$ $P(A \cup B)=P(A)+P(B)$, As the events are independent...

We are having two events $A$ and $B$ such that $\begin{array}{l} P(A)=\frac{1}{2}, P(B)=\frac{7}{12} \text { and } P\left(A^{\prime} \cup B^{\prime}\right)=\frac{1}{4} \\ P\left(A^{\prime} \cup...

$\begin{array}{l} P(A)=\frac{6}{11}, P(B)=\frac{5}{11} \text { and } P(A \cup B)=\frac{7}{11} \\ P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ \Rightarrow \frac{7}{11}=\frac{6}{11}+\frac{5}{11}-P(A \cap B)...

$\begin{array}{l} P(B / A)=\frac{P(A \cap B)}{P(A)} \\ \Rightarrow \text { And } P(A)=0.8 \\ \Rightarrow P(A \cap B)=0.32 \end{array}$ So, $P(A / B)=\frac{P(A \cap B)}{P(B)}$ $\Rightarrow P(A /...

As $A$ and $B$ are independent events such that $P(A)=0.4, P(B)=x$ So, $P(A \cap B)=P(A) P(B)$ And $P(A \cup B)=P(A)+P(B)+P(A \cap B)$ $\begin{array}{l} P(A \cup B)=0.4+X-0.4 X=0.5 \\ \Rightarrow...

If $A$ and $B$ are mutually exclusive events then, $P(A)=0.4, P(B)=X$ And $P(A \cup B)=P(A)+P(B)=0.5=0.4+P(B)$ $\Rightarrow P(B)=0.1$

Variance can not be greater than mean as then, q wll be greater than 1, which is not possible. As, $n p=6$ and $n p q=9$ $q=3 / 2 \ldots$ (not possible)

Mean is 10, Standard deviation is $2 \sqrt{2}$ So, variance is $\sigma^{2}$ i.e. 8 Thus, Mean $=\mathrm{np}=10$ Variance $=\mathrm{npq}=8$ $\Rightarrow \mathrm{q}=\frac{4}{5}$ $\Rightarrow...

$\begin{array}{l} \text { Mean }+\text { Variance }=n p+n p q=n p(1+q)=25 / 3 \\ \text { Variance }=n^{2} p^{2} q=n 2=50 / 3 \ldots(i) \\ n^{2} p^{2}(1+q)^{2}=625 / 9 \ldots \text { (ii) }...

$\begin{array}{l} \text { Mean }=\mathrm{np}=6 \\ \text { Variance }=\mathrm{npq}=2 \\ \Rightarrow \mathrm{q}=\frac{1}{3} \\ \Rightarrow \mathrm{p}=1-\frac{1}{3}=\frac{2}{3} \\ \Rightarrow...

$\begin{array}{l} \text { Mean }=n p=4 \\ \text { Variance }=n p q=4 / 3 \\ \Rightarrow q=\frac{1}{3} \\ \Rightarrow p=1-\frac{1}{3}=\frac{2}{3} \\ \Rightarrow n=6 \end{array}$ The probability $(X...

$\begin{array}{l} \text { Mean }=n p=5 \\ \text { Variance }=n p q=2.5 \\ \Rightarrow q=\frac{2.5}{5}=\frac{1}{2} \\ \Rightarrow p=1-\frac{1}{2}=\frac{1}{2} \\ \Rightarrow n=10 \end{array}$...

$\begin{array}{l} \text { Mean }=\mathrm{np}=9 \\ \text { Variance }=\mathrm{npq}=6 \\ \Rightarrow \mathrm{q}=\frac{6}{9}=\frac{2}{3} \\ \Rightarrow \mathrm{p}=1-\frac{6}{9}=\frac{1}{3} \\...

Probability of getting an even number is $=3 / 6=1 / 2$ Probability of getting an odd number is $=3 / 6=1 / 2$ $\begin{array}{l} \text { Variance }=\mathrm{npq} \\ \Rightarrow 100 \times \frac{1}{2}...

Using Bernoulli's Trial we have, $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . n$ and $q=(1-p), n=3$ $\begin{array}{l} p=2 / 6=1 / 3, q=4 / 6=2 / 3 \\...

The probability that the burglar will be hit by a bullet is $0.6 .$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2,...

Balls are drawn at random, So, the probability that at least one is white is, In a trial the probability of selecting a white ball is $\frac{5}{20}$ So, in 4 trials the probability that at least one...

(i) Balls are drawn at random, So, the probability that none is white is, In a trial the probability of selecting a non-white ball is $\frac{15}{20}$ So, in 4 trials it will be,...

The probability that a man aged 60 will live to be 70 is $0.65$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots...

The probability that the bird will be shot, is $1 / 3$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots \ldots ....

The probability that the hurdle will be cleared is $5 / 6$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots...

Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots \ldots . n \text { and } q=(1-p), n=7 \\ p= q=? \end{array}$...

The probability that the operations performed are successful is $=0.8$ The probability that at least three operations are successful is $=\mathrm{P}(3)+\mathrm{P}(4)$ $\begin{array}{l} \Rightarrow{...

The probability that any one of them has an accident is $0.1$. The probability any car reaches safely is $0.9 .$ The probability that all the cars reach the finishing line without any accident is...

The probability that the called number is busy is $\frac{1}{15}$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2,...

The probability that the item is defective $=\frac{1}{10}=p$ The probability that the bulb will not fuse $=1-\frac{1}{10}=\frac{9}{10}=q$ Using Bernoulli's we have, $\begin{array}{l} P(\text {...

(i) The probability that the item is defective $=\frac{1}{10}=p$ The probability that the bulb will not fuse $=1-\frac{1}{10}=\frac{9}{10}=\mathrm{q}$ Using Bernoulli's we have, $\begin{array}{l}...

The probability that the bulb will fuse $=0.05=p$ The probability that the bulb will not fuse $=1-0.05=0.95=q$ Using Bernoulli's we have, $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$...

(i) The probability that the bulb will fuse $=0.05=\mathrm{p}$ The probability that the bulb will not fuse $=1-0.05=0.95=q$ Using Bernoulli's we have, $\begin{array}{l} P(\text { Success }=x)={...

(i) Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n \text { and } q=(1-p), n=5$ The probability of success, i.e. the bulb is...

Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . .$ and $q=(1-p), n=8$ The probability of success, i.e. the bulb is defective...

(i) Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots \ldots$ and $q=(1-p), n=7$ the favourable outcomes,...

(i) Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p), n=7$ the favourable outcomes,...

As the pair of die is thrown 4 times, The total number of outcomes $=36$ Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and...

The total outcomes $=36$ The favourable outcomes are $(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(4,1),(4,2),(4,3),(4,5),(4,6)$ Thus, the probability $=$ favourable outcomes/total outcomes $\Rightarrow...

Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p)$ We know that the favourable outcomes of getting at most 2 successes...

(i) Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots \ldots \text { and } q=(1-p)$ We know that the favourable outcomes of getting exactly...

Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p)$ As the die is thrown 6 times the total number of outcomes will be...

(i) Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . . n$ and $q=(1-p)$ As the die is thrown 6 times the total number of outcomes will...

As 10 coins are tossed simultaneously the total number of outcomes are $2^{10}=1024$. the favourable outcomes of getting at least 4 heads will be ${ }^{10} \mathrm{C}_{4}+{ }^{10} \mathrm{C}_{5}+{...

(i) As 10 coins are tossed simultaneously the total number of outcomes are $2^{10}=1024$. the favourable outcomes of getting exactly 3 heads will be ${ }^{10} \mathrm{C}_{3}=120$ Thus, the...

As the coin is tossed 6 times the total number of outcomes will be $2^{6}=64$ And we know that the favourable outcomes of getting at most 4 heads will be ${ }^{6} \mathrm{C}_{0}+{ }^{6}...

(i) As the coin is tossed 6 times the total number of outcomes will be $2^{6}=64$ And we know that the favourable outcomes of getting exactly 4 heads will be ${ }^{6} c_{4}=15$ Thus, the probability...

As 7 coins are tossed simultaneously the total number of outcomes are $2^{7}=128$. The favourable number of outcomes that a tail appears an odd number of times will be, ${ }^{7} \mathrm{C}_{1}+{...

As the coin is tossed 5 times the total number of outcomes will be $2^{5}=32$. And we know that the favourable outcomes of a head appearing even number of times will be, That either the head appears...

As the coin is tossed 6 times the total number of outcomes will be $2^{6}$. And we know that the favourable outcomes of getting at least 3 heads will be ${ }^{6} c_{3}+{ }^{6} c_{4}+{ }^{6} c_{5}+{...