Question

Find the probability of 2 or 3 successes in 7 trials of a binomial experiment in which the probability of success in any one trial is 42%. Round to the nearest tenth of a percent.

Answer 1

53.7 %

Explanation:

This is the answer on acellus

Acellus sux

Answer 2

53.7 %

Explanation:

Given:

Number of trials,
`n=7`

Number of successes,
`x=2,3`

Probability of success,
`p=42 \%= (42)/(100)=0.42`

Therefore, probability of failure,
`q=1-p=1-0.42=0.58`

The Bernoulli's distribution for
`x` successes out of
`n` trials is given as:

`P(X=x)=_(x)^(n)\textrm{C}p^(x)q^((n-x))`

So, probability of 2 or 3 successes is given as:

`P(X=2\textrm{ or}X=3)=P(X=2)+P(X=3)\\\\P(X=2\textrm{ or}X=3)=_(2)^(7)\textrm{C}(0.42)^(2)(0.58)^((7-2))+_(3)^(7)\textrm{C}(0.42)^(3)(0.58)^((7-3))\\\\P(X=2\textrm{ or}X=3)=21* (0.42)^(2)(0.58)^(5)+35* (0.42)^(3)(0.58)^(4)\\\\P(X=2\textrm{ or}X=3)= 0.5365=53.65 \% \approx 53.7 \%`

Therefore, the probability of 2 or 3 successes in 7 trials of a binomial experiment is 53.7 %.