question_answer

In Arjun's school, annual sports meet with the title 'Sports for Healthy Life' was being organised. On the last day of the meet, there was the event of hurdle race. In this hurdle, it was decided that a player has to cross 10 hurdles.

From the past games experience it can be said that the probability of a player to clear each hurdle will be 5/6. Find the probability that a player will knock down fewer than 2 hurdles.

OR

A couple has 2 children. Find the probability that both are boys, if it is known that

(i) One of them is a boy,

(ii) The older child is a boy.

Answer:

Given, a player has to cross 10 hurdles, so n =10.

Also, probability of a player to clear each hurdle, \[q=\frac{5}{6}\]

Then, probability of a player to not clear a hurdle,

\[p=1-\frac{5}{6}=\frac{1}{6}\]             \[[\because p\,+q=1]\]

Let X denotes the number of hurdles knock down.

Then, X can take values 0, 1, 2,..., 10

Now, the required probability = Probability that a player will knock down fewer than 2 hurdles

\[P(X=0)+P(X=1)\]

\[{{=}^{10}}{{C}_{0}}{{\left( \frac{1}{6} \right)}^{0}}{{\left( \frac{5}{6} \right)}^{10}}{{+}^{10}}{{C}_{1}}{{\left( \frac{1}{6} \right)}^{1}}{{\left( \frac{5}{6} \right)}^{9}}\]

\[[\because P(X=r){{=}^{n}}{{C}_{r}}\,{{p}^{r}}{{q}^{n\,-\,r}}]\]

\[=1{{\left( \frac{5}{6} \right)}^{10}}+10\left( \frac{{{5}^{9}}}{{{6}^{10}}} \right)\]

\[=\frac{{{5}^{9}}}{{{6}^{10}}}(5\,+10)=\frac{{{5}^{9}}}{{{6}^{9}}}\times \frac{15}{6}\]

\[=\frac{{{5}^{9}}}{{{6}^{9}}}\times \frac{5}{2}=\frac{{{5}^{10}}}{2\times {{6}^{9}}}\]

OR

Let B represents the boy and G represents the girl. Then, the sample space of given problem is

\[S=\{BB,\,GB,\,BG,\,GG\}\]

\[\therefore \]      \[n(S)=4\]

Now, let E be the event that both children are boys, i.e.

\[E=\{BB\}\]

(i) Let \[{{F}_{1}}\] be the event that one of the children is boy.

Then, \[{{F}_{1}}=\{BB,\,\,GB,\,\,BG\}\]

\[\therefore \] Required probability

\[=P(E/{{F}_{1}})=\frac{P(E\cap {{F}_{1}})}{P({{F}_{1}})}=\frac{P(BB)}{P({{F}_{1}})}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\]

(ii) Let \[{{F}_{2}}\] be the event that the older child is a boy. Then, \[{{F}_{2}}=\{BB,\,\,BG\}\]

\[\therefore \] Required probability \[=P(E\,/{{F}_{2}})=\frac{P(E\,\cap {{F}_{2}})}{P({{F}_{2}})}\]

\[=\frac{P(BB)}{P({{F}_{2}})}=\frac{1/4}{2/4}=\frac{1}{2}\]