Question

In the binomial probability distribution, let the number of trials be n = 3, and let the probability of success be p = "0.3229." Use a calculator to compute the following.
(a) The probability of two successes. (Round your answer to three decimal places.)
(b) The probability of three successes. (Round your answer to three decimal places.)
(c) The probability of two or three successes.

Answer 1

Answer:

a) So, the probability is P=0.212.

b) So, the probability is P=0.034.

c) So, the probability is P=0.239.

Step-by-step explanation:

We know that the number of trials be n = 3, and let the probability of success be p = "0.3229." We get that q=1-0.3229=0.6771.

We use the formula:

$\boxed{P(X=k)=C_k^n\cdot p^k\cdot q^{n-k}}$

a) We calculate the probability of two successes, so k=2.

$P(X=2)=C_2^3\cdot 0.3229^2\cdot 0.6771^1\\\\P(X=2)=3\cdot 0.0706\\\\P(X=2)=0.212$

So, the probability is P=0.212.

b) We calculate the probability of three successes, so k=3.

$P(X=3)=C_3^3\cdot 0.3229^3\cdot 0.6771^0\\\\P(X=3)=1\cdot 0.034\cdot 1\\\\P(X=3)=0.034$

So, the probability is P=0.034.

c) We calculate the probability of two or three successes.

$P(2\cup3)=P(2)+P(3)-P(2\cap 3)\\\\P(2\cup3)=0.212+0.034-P(2)\cdot P(3)\\\\P(2\cup3)=0.246-0.007\\\\P(2\cup3)=0.239$

So, the probability  is P=0.239.