QUESTION

# If a sex ratio of births is 49 girls to 51 boys, the probability that there will be 8 girls amongst 10 babies born on the same day in a maternity hospital is,(a) $^{10}{{C}_{8}}{{\left( 0.51 \right)}^{8}}{{\left( 0.49 \right)}^{2}}$ (b) $^{10}{{C}_{8}}{{\left( 0.49 \right)}^{8}}{{\left( 0.51 \right)}^{2}}$(c) $^{10}{{C}_{8}}{{\left( 0.49\times 0.51 \right)}^{8}}$ (d) $^{10}{{C}_{8}}{{\left( 0.49\times 0.51 \right)}^{2}}$

Hint: In this question, we will first find the probability of birth of a girl and boy. Then we will use the formula of finding probability of occurrence of E $r$ times out of $n$ times, to find the answer.

Given that, the ratio between the number of births of girls to the number of births of boys is: 49:51.

Therefore, among 49+51=100 children born, 49 are girls and 51 are boys.

Also, probability of occurrence of an event E, represented by P(E), is given by,

$\text{P}\left( \text{E} \right)\text{=}\dfrac{\text{n}\left( \text{E} \right)}{\text{n}\left( \text{S} \right)}\cdots \cdots \left( i \right)$

Where, n(E) is the number of favourable outcomes of E,

And n(S) is the total number of events in the sample space.

Let the event of the birth of girl child be G, and that of boy child be B.

Then, we have,

n(G) = 49, n(B) = 51, n(S) = 100.

Then, using these values in probability formula, we get,

$\text{P}\left( \text{G} \right)\text{=}\dfrac{\text{n}\left( \text{G} \right)}{\text{n}\left( \text{S} \right)}$

\begin{align} & =\dfrac{49}{100} \\ & =0.49 \\ \end{align}

And, $\text{P}\left( B \right)\text{=}\dfrac{\text{n}\left( B \right)}{\text{n}\left( B \right)}$

\begin{align} & =\dfrac{51}{100} \\ & =0.51 \\ \end{align}

Therefore, probability of having a girl = 0.49.

And, probability of having a boy = 0.51.

Now, we know, number of ways in which $r$ things can be selected out of $n$ things is given by $^{n}{{C}_{r}}$.

Therefore, the number of ways in which 8 girls can be selected out of 10 babies is $^{10}{{C}_{8}}$.

Also, out of $n$ times, probability of occurrence of E $r$times is given by : $^{n}{{C}_{r}}\times \text{P}{{\left( \text{E} \right)}^{\text{r}}}\text{ }\!\!\times\!\!\text{ P}{{\left( {{\text{E}}^{\text{ }\!\!'\!\!\text{ }}} \right)}^{\text{n-r}}}$, where, $\text{P}\left( {{\text{E}}^{\text{ }\!\!'\!\!\text{ }}} \right)$ is probability of not occurrence of E.

Here, for the event of birth of a girl child, that is event G, probability of not occurrence of G is occurrence of B, that is birth of boy child. Therefore, $\text{P}\left( {{\text{E}}^{\text{ }\!\!'\!\!\text{ }}} \right)\text{=P}\left( \text{B} \right)\text{=0}\text{.51}$.

Therefore, probability that there will be 8 girls amongst 10 babies born is given by,

$^{n}{{C}_{r}}\times \text{P}{{\left( \text{E} \right)}^{\text{r}}}\text{ }\!\!\times\!\!\text{ P}{{\left( {{\text{E}}^{\text{ }\!\!'\!\!\text{ }}} \right)}^{\text{n-r}}}$

\begin{align} & {{=}^{10}}{{C}_{8}}\text{ }\!\!\times\!\!\text{ P}{{\left( \text{G} \right)}^{\text{8}}}\text{ }\!\!\times\!\!\text{ P}{{\left( \text{B} \right)}^{\text{10-8}}} \\ & {{=}^{10}}{{C}_{8}}\text{ }\!\!\times\!\!\text{ P}{{\left( \text{G} \right)}^{\text{8}}}\text{ }\!\!\times\!\!\text{ P}{{\left( \text{B} \right)}^{2}} \\ & {{=}^{10}}{{C}_{8}}\text{ }\!\!\times\!\!\text{ }{{\left( 0.49 \right)}^{\text{8}}}\text{ }\!\!\times\!\!\text{ }{{\left( 0.51 \right)}^{2}} \\ \end{align}

Hence, the correct answer is option (b).

Note: In this question, we can also first find the probability that there will be two boys out of 10 babies with the same ratio data and then use it to find the answer by subtracting it from one.