Respuesta :
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 [tex]\frac{1}{2}[/tex]
The probability of getting 6 girl is [tex](\frac{1}{2})^6[/tex]
Since we have 6 girls,
We also need to find the probability of getting 2 boys, which is
[tex](\frac{1}{2})^2[/tex]
There are [tex]^8C_6[/tex] ways to get 6 girls in 8 births
i.e, [tex]^8C_6=\frac{8!}{6!\times 2!}[/tex]
[tex]^8C_6=\frac{8\times 7\times 6!}{6!\times 2\times 1}[/tex]
[tex]^8C_6=28[/tex]
There are 28 ways to get 6 girls in 8 births.
The probability of getting exactly 6 girls in 8 births.
[tex]=28\times (\frac{1}{2})^6\times (\frac{1}{2})^2[/tex]
[tex]=28\times 0.015625\times 0.25[/tex]
[tex]=0.109375[/tex]
Therefore, The probability of getting exactly 6 girls in 8 births is 0.1093
The probability of having exactly [tex]6[/tex] girls in [tex]8[/tex] birth is [tex]\boxed{\bf 0.10937}[/tex].
Further explanation:
Concept used:
The probability of an event [tex]E[/tex] is calculated as follows:
[tex]\boxed{P(E)=\dfrac{n(E)}{n(S)}}[/tex]
Here, [tex]n(E)[/tex] is the number of favorable outcomes in an event [tex]E[/tex] and [tex]n(S)[/tex] is the number of element in sample space [tex]S[/tex].
The probability of exactly [tex]r[/tex] success in [tex]n[/tex] trial is expressed as follows:
[tex]\boxed{P(F)=^{n}C_{r}p^{r}q^{n-r}}[/tex]
Here, [tex]r[/tex] is the probability of success in an event and [tex]q[/tex] is the probability of failure.
Calculation:
Consider [tex]A[/tex] as an event of having a girl in first birth and [tex]P(A)[/tex] as the probability of event [tex]A[/tex].
The probability of having a girl is [tex]P(A)=\frac{1}{2}[/tex].
Assume that [tex]A'[/tex] is the complement event of an event [tex]A[/tex] and [tex]P(A')[/tex] is the probability of complementary event [tex]A'[/tex].
The event is the event of not having a girl in the first birth.
The probability of event [tex]A'[/tex] is calculated as follows:
[tex]\boxed{P(A')=1-P(A)}[/tex] …… (1)
Substitute [tex]P(A)=\frac{1}{2}[/tex] in the equation (1) to obtain the probability of event [tex]A'[/tex].
[tex]\begin{aligned}P(A')&=1-\dfrac{1}{2}\\&=\dfrac{1}{2}\end{aligned}[/tex]
The probability of having exactly [tex]6[/tex] girls in [tex]8[/tex] birth is calculated as follows:
[tex]\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}[/tex]
Thus, the probability of having exactly [tex]6[/tex] girls in [tex]8[/tex] birth is [tex]\boxed{\bf 0.10937}[/tex].
Learn more:
1. Learn more about problem on numbers: https://brainly.com/question/1852063
2. Learn more about problem on function https://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).
