Chapter 13: Probability

Ans: It is given in the question that $P\left( A \right)\ne 0$ and A is a subset of B

$\therefore P\left( A\bigcap B \right)=P\left( A \right)$

$\therefore P\left( B\left| A \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( A \right)}$

$\Rightarrow P\left( B\left| A \right. \right)=\frac{P\left( A \right)}{P\left( A \right)}=1$

Ans: since it is given that $A\bigcap B=\phi $

Therefore $\text{             }P\left( A\bigcap B \right)=0$

$\therefore P\left( B\left| A \right. \right)=\frac{P\left( A\bigcap B \right)}{P\left( A \right)}=0$

Ans: The sample space for a family to have two children is given by

$S=\left\{ BB,GG,BG,GB \right\}$

Let us have following notations and their probabilities given as shown

N: Both children are males

$P\left( N \right)=\frac{1}{4}$

K: at least one of the children is male

\[P\left( K \right)=\frac{3}{4}\]

$\because N\bigcap K=BB$

$\therefore P\left( N\bigcap K \right)=\frac{1}{4}$

Now probability that both of the children are males provided at least one of the child is male is given by

$P\left( N\left| K \right. \right)=\frac{P\left( N\bigcap K \right)}{P\left( K \right)}$ 

$P\left( N\left| K \right. \right)=\frac{\frac{1}{4}}{\frac{3}{4}}$

Hence probability that both of the children are males provided at least one of the child is male is $\frac{1}{3}$

Ans: The sample space for a family to have two children is given by

$S=\left\{ BB,GG,BG,GB \right\}$

Let us have following notations and their probabilities given as shown

N: Both children are females

$P\left( N \right)=\frac{1}{4}$

K: elder child is a female

\[P\left( K \right)=\frac{2}{4}\]

$\because N\bigcap K=GG$

$\therefore P\left( N\bigcap K \right)=\frac{1}{4}$

Now probability that both of the children are males provided atleast one of the child is male is given by

$P\left( N\left| K \right. \right)=\frac{P\left( N\bigcap K \right)}{P\left( K \right)}$ 

$P\left( N\left| K \right. \right)=\frac{\frac{1}{4}}{\frac{2}{4}}$

Hence probability that both of the children are males provided atleast one of the child is male is $\frac{1}{2}$

Ans: It is given that $5$percent of males and $0.25$ percent of females have grey hair

Therefore total people having grey hair is $\left( 5+0.25 \right)=5.25$ percent 

Hence probability of male being haired is $\frac{5}{5.25}$

Ans: Given probability of a person being right-handed is$p=0.9$  

Since a person can only be right-handed or left handed 

Therefore it follows binomial distribution with

$n=10,p=\frac{9}{10},q=\frac{1}{10}$

Probability that at least $6$ people are right handed is given by

\[{{\sum\limits_{k=0}^{k=7}{^{10}{{C}_{k}}\left( \frac{9}{10} \right)}}^{k}}{{\left( \frac{1}{10} \right)}^{10-k}}\]

Therefore probability that at most $6$ people are right handed is given by

\[1-{{\sum\limits_{k=0}^{k=7}{^{10}{{C}_{k}}\left( \frac{9}{10} \right)}}^{k}}{{\left( \frac{1}{10} \right)}^{10-k}}\]

Ans: from the given data we have

Probability of balls bearing X mark is $p=\frac{10}{25}$

 Probability of balls bearing Y mark is $q=\frac{15}{25}$

Hence we can say that random variable K which measures the number of Y marked balls follows binomial distribution with

$n=6,p=\frac{2}{5},q=\frac{3}{5}$

Probability that all ball bear X is same as none of balls bear Y which is given by

$P\left( K=0 \right){{=}^{6}}{{C}_{0}}{{\left( \frac{3}{5} \right)}^{0}}{{\left( \frac{2}{5} \right)}^{6}}$

Hence Probability that all ball bear X is ${{\left( \frac{2}{5} \right)}^{6}}$

Ans: Probability of not more than $2$MY balls are selected is given by

$\text{      }P\left( k\le 2 \right)=P\left( K=0 \right)+P\left( K=1 \right)+P\left( K=2 \right)$

$\Rightarrow P\left( k\le 2 \right){{=}^{6}}{{C}_{0}}{{\left( \frac{3}{5} \right)}^{0}}{{\left( \frac{2}{5} \right)}^{6}}{{+}^{6}}{{C}_{1}}{{\left( \frac{3}{5} \right)}^{1}}{{\left( \frac{2}{5} \right)}^{5}}{{+}^{6}}{{C}_{2}}{{\left( \frac{3}{5} \right)}^{2}}{{\left( \frac{2}{5} \right)}^{4}}$

$\Rightarrow P\left( k\le 2 \right)={{\left( \frac{2}{5} \right)}^{4}}\left( \frac{4}{25}+6{{\left( \frac{3}{5} \right)}^{1}}\left( \frac{2}{5} \right)+15{{\left( \frac{3}{5} \right)}^{2}} \right)$

Hence Probability of not more than $2$ balls are selected is $7{{\left( \frac{2}{5} \right)}^{4}}$

Ans: Probability of at least one ball bear Y is given by

$P\left( K\ge 1 \right)=1-P\left( k=0 \right)$

$\Rightarrow P\left( K\ge 1 \right)=1-{{\left( \frac{2}{5} \right)}^{6}}$

Hence Probability of at least one ball bear Y is $1-{{\left( \frac{2}{5} \right)}^{6}}$

Ans: Since there were $6$ balls therefore each are of $3$ if equal 

Probability that number of X and Y is equal is given by

\[P\left( K=3 \right){{=}^{6}}{{C}_{3}}{{\left( \frac{2}{5} \right)}^{3}}{{\left( \frac{3}{5} \right)}^{3}}\]

 Hence  Probability that number of X and Y is equal is $\frac{834}{3125}$

Ans: Given probability that he will clear each hurdle is $p=\frac{5}{6}$ 

And hence probability that he knocks down is given as $q=\frac{1}{6}$

Let k be the random variable that represents the hurdles he knocks down 

Clearly K follows binomial distribution with

$n=10,p=\frac{5}{6},q=\frac{1}{6}$

Probability that he knocks down less than $2$hurdles is given by

$P\left( K=0 \right)+P\left( K=1 \right)$

$^{10}{{C}_{0}}{{\left( \frac{1}{6} \right)}^{0}}{{\left( \frac{5}{6} \right)}^{10}}{{+}^{10}}{{C}_{1}}{{\left( \frac{1}{6} \right)}^{1}}{{\left( \frac{5}{6} \right)}^{9}}=\frac{{{5}^{10}}}{{{6}^{9}}\times 2}$

Hence Probability that he knocks down less than $2$hurdles is $\frac{{{5}^{10}}}{{{6}^{9}}\times 2}$

Ans: From the question we have 

Probability of getting a six is $\text{ }p=\frac{1}{6}$

Probability of getting no six is $q=\frac{5}{6}$

Let K be the random variable that represents the number of sixes 

Clearly K follows binomial distribution with

$n=5,p=\frac{1}{6},q=\frac{5}{6}$

Probability that two sixes come in first two throws is given by

 $P\left( K=2 \right){{=}^{5}}{{C}_{2}}{{\left( \frac{1}{6} \right)}^{2}}{{\left( \frac{5}{6} \right)}^{3}}$

Also Probability that third sixe come is given by

$\frac{1}{6}P\left( K=2 \right)$

Hence probability of getting three sixes is $\frac{10\times 125}{{{6}^{6}}}$

Ans: In a leap year, we have $366$ days i.e., $52$weeks and $2$ days. 

In $52$ weeks, we have $52$ Tuesdays.

Therefore, the probability that the leap year will contain 53 Tuesdays is equal to the probability that the remaining $2$ days will be Tuesdays. 

The remaining $2$days can be any of the following 

Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday and Sunday and Monday 

Total number of cases = 7

⸫ Probability that a leap year will have $53$ Tuesdays $=\frac{2}{7}$

Ans: Given that the probability of success is twice that of failure

$\therefore p=2q$

Also we know that sum of probabilities of exhaustive events equals $1$

$\therefore p=\frac{2}{3},q=\frac{1}{3}$

Let K be the random variable that represents successes 

$\therefore P\left( K\ge 4 \right)=P\left( K=4 \right)+P\left( K=5 \right)+P\left( K=6 \right)$

$\Rightarrow P\left( K\ge 4 \right){{=}^{6}}{{C}_{4}}{{\left( \frac{2}{3} \right)}^{4}}{{\left( \frac{1}{3} \right)}^{2}}{{+}^{6}}{{C}_{3}}{{\left( \frac{2}{3} \right)}^{3}}{{\left( \frac{1}{3} \right)}^{1}}{{+}^{6}}{{C}_{6}}{{\left( \frac{2}{3} \right)}^{6}}{{\left( \frac{1}{3} \right)}^{0}}$

Thus on solving we found the probability of getting at least $4$ successes in six trials is $\frac{31}{9}{{\left( \frac{2}{3} \right)}^{4}}$

Ans: Let a person tosses the coin n times , n being Bernoulli Trials 

Here probability of head equals that of tails and each equal to $\frac{1}{2}$ 

Numbers of trials for which at least one head is more than $0.9$ is given by

$1-P\left( K=0 \right)>0.9$

$\Rightarrow 1{{-}^{n}}{{C}_{0}}{{\left( \frac{1}{2} \right)}^{n}}>\frac{9}{10}$

$\Rightarrow {{2}^{n}}>10$

Hence the person must toss more than $4$ times