question_answer

The probabilities of two students A and B coming to the school in time are  \[\frac{3}{7}\] and \[\frac{5}{7},\] respectively. Assuming that the events, ?A coming in time' and 'B coming in time' are independent, find the probability, of only one of them coming to the school in time

OR

Find the probability that in 10 throws of a fair die, a score which is a multiple of 3 will be obtained in at least 8 of the throws.

Answer:

P(A) = Probability of student A coming to school in time \[=\frac{3}{7}\]

P(B) = Probability of student B coming to school in time \[=\frac{5}{7}\]

Then, the probability that only one of students coming to school in time

\[=P(A\,\cap \bar{B})+P(\bar{A}\,\cap B)\]

\[=P(A)\times P(\bar{B})+P(\bar{A})\times P(B)\]

\[=P(A)\times [1-P(B)]+[1-P(A)]\times P(B)\]

\[=\frac{3}{7}\times \left( 1-\frac{5}{7} \right)+\left( 1-\frac{3}{7} \right)\times \frac{5}{7}\]  \[=\frac{3}{7}\times \frac{2}{7}+\frac{4}{7}\times \frac{5}{7}\]

\[=\frac{6}{49}+\frac{20}{49}=\frac{26}{49}\]

OR

Here, success is a score which is a multiple of 3, i.e. 3 or 6.

Let p = probability of success \[=\frac{2}{6}=\frac{1}{3}\]

and q = probability of failure \[=1-\frac{1}{3}=\frac{2}{3}\]

Now, let X be a random variable that denotes the number of success in 10 throws of a fair die. Then, X can take values 0, 1, 2,...,10.

Clearly, X follows binomial distribution with n = 10 and \[p=\frac{1}{3}.\]

\[\therefore P(X=x){{=}^{10}}{{C}_{x}}{{\left( \frac{1}{3} \right)}^{x}}{{\left( \frac{2}{3} \right)}^{10-x}};x=0,\,\,1,\,\,2,...10\]

Now, required probability \[=P(X\ge 8)\]

\[=P(X=8)+P(X=9)+P(X=10)\]

\[{{=}^{10}}{{C}_{8}}{{\left( \frac{1}{3} \right)}^{8}}{{\left( \frac{2}{3} \right)}^{2}}{{+}^{10}}{{C}_{9}}{{\left( \frac{1}{3} \right)}^{9}}\left( \frac{2}{3} \right){{+}^{10}}{{C}_{10}}{{\left( \frac{1}{3} \right)}^{10}}\]

\[={{\left( \frac{1}{3} \right)}^{10}}{{[}^{10}}{{C}_{2}}.4\,{{+}^{10}}{{C}_{1}}\cdot 2+1]\]

\[={{\left( \frac{1}{3} \right)}^{10}}[45\times 4+20+1]=\frac{201}{{{3}^{10}}}\]