Question

Find the probability of getting exactly 6 girls in 8 births.

Answer 1

Answer:

The probability of getting exactly 6 girls in 8 births is 0.1093

Step-by-step explanation:

Given : 6 girls in 8 births.

To find : The probability of getting exactly 6 girls in 8 births.

Solution :

The probability of getting girl is $\frac{1}{2}$

The probability of getting 6 girl is $(\frac{1}{2})^6$

Since we have 6 girls,

We also need to find the probability of getting 2 boys, which is  

$(\frac{1}{2})^2$

There are $^8C_6$ ways to get 6 girls in 8 births

i.e, $^8C_6=\frac{8!}{6!\times 2!}$

$^8C_6=\frac{8\times 7\times 6!}{6!\times 2\times 1}$

$^8C_6=28$

There are 28 ways to get 6 girls in 8 births.

The probability of getting exactly 6 girls in 8 births.

$=28\times (\frac{1}{2})^6\times (\frac{1}{2})^2$

$=28\times 0.015625\times 0.25$

$=0.109375$

Therefore, The probability of getting exactly 6 girls in 8 births is 0.1093

Answer 2

The probability of having exactly $6$ girls in $8$ birth is $\boxed{\bf 0.10937}$.

Further explanation:

Concept used:

The probability of an event $E$ is calculated as follows:

$\boxed{P(E)=\dfrac{n(E)}{n(S)}}$  

Here, $n(E)$ is the number of favorable outcomes in an event $E$ and $n(S)$ is the number of element in sample space $S$.

The probability of exactly $r$ success in $n$ trial is expressed as follows:

$\boxed{P(F)=^{n}C_{r}p^{r}q^{n-r}}$

Here, $r$ is the probability of success in an event and $q$ is the probability of failure.

Calculation:

Consider $A$ as an event of having a girl in first birth and $P(A)$ as the probability of event $A$.

The probability of having a girl is $P(A)=\frac{1}{2}$.

Assume that $A'$ is the complement event of an event $A$ and $P(A')$ is the probability of complementary event $A'$.

The event is the event of not having a girl in the first birth.

The probability of event $A'$ is calculated as follows:

$\boxed{P(A')=1-P(A)}$   …… (1)

Substitute $P(A)=\frac{1}{2}$ in the equation (1) to obtain the probability of event $A'$.

$\begin{aligned}P(A')&=1-\dfrac{1}{2}\\&=\dfrac{1}{2}\end{aligned}$

 

The probability of having exactly $6$ girls in $8$ birth is calculated as follows:

$\begin{aligned}P(F&)=^{8}C_{6}\left(\dfrac{1}{2}\right)^{6}\left(\dfrac{1}{2}\right)^{8-6}\\&=\dfrac{8!}{6!\cdot 2!}\cdot (0.5)^{6}\cdot (0.5)^{2}\\&=28(0.5)^{8}\\&=0.10937\end{aligned}$  

Thus, the probability of having exactly $6$ girls in $8$ birth is $\boxed{\bf 0.10937}$.

Learn more:

1. Learn more about problem on numbers: brainly.com/question/1852063

2. Learn more about problem on function brainly.com/question/3225044

Answer details:

Grade: Senior school

Subject: Mathematics

Chapter: Probability

Keywords: Probability, exact event, sample space, number of element, complement event, success, failure, favorable, trial, P(E)=n(E)/n(S), P(A')=1-P(A).