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:
[tex]\boxed{P(X=k)=C_k^n\cdot p^k\cdot q^{n-k}}[/tex]
a) We calculate the probability of two successes, so k=2.
[tex]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\\[/tex]
So, the probability is P=0.212.
b) We calculate the probability of three successes, so k=3.
[tex]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\\[/tex]
So, the probability is P=0.034.
c) We calculate the probability of two or three successes.
[tex]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[/tex]
So, the probability is P=0.239.
