Answer:
Hence, the probability of exactly 3 successes in 6 trials of a binomial experiment round to the nearest tenth of a percent is:
31.2%
Step-by-step explanation:
The probability of getting exactly k successes in n trials is given by the probability mass function:
[tex]{\displaystyle P(k;n,p)=P(X=k)={\binom {n}{k}}p^{k}(1-p)^{n-k}}[/tex]
Where p denotes the probability of success.
We are given that the probability of success if 50%.
i.e. [tex]p=\dfrac{1}{2}[/tex]
also form the question we have:
k=3 and n=6.
Hence the probability of exactly 3 successes in 6 trials is:
[tex]{\displaystyle P(3;6,\dfrac{1}{2})=P(X=3)={\binom {6}{3}}(\dfrac{1}{2})^{3}(1-\dfrac{1}{2})^{6-3}}[/tex]
[tex]{\displaystyle P(3;6,\dfrac{1}{2})=P(X=3)={\binom {6}{3}}(\dfrac{1}{2})^{3}(\dfrac{1}{2})^{3}}[/tex]
[tex]{\displaystyle P(3;6,\dfrac{1}{2})=P(X=3)={\binom {6}{3}}(\dfrac{1}{2})^{6}[/tex]
[tex]\binom {6}{3}=20[/tex]
Hence,
[tex]{\displaystyle P(3;6,\dfrac{1}{2})=P(X=3)=20\times (\dfrac{1}{2})^6=\dfrac{5}{16}[/tex]
In percentage the probability will be:
[tex]\dfrac{5}{16}\times 100=31.25\%=31.2\%[/tex]