A rational number can be written as the ratio of on BLANK to another.

The prior probabilities for events A1 and A2 are P(A1) = 0.30 and P(A2) = 0.55. It is also known that P(A1 ∩ A2) = 0. Suppose P(

k0ka [10]

We're given the following probabilities:

$P(A_1)=0.30$

$P(A_2)=0.55$

$P(A_1\cap A_2)=0$

$P(B\mid A_1)=0.20$

$P(B\mid A_2)=0.05$

(a) Yes, $A_1$ and $A_2$ are mutually exclusive. This is exactly what zero probability of their intersection means. The two events cannot occur simultaneously.

(b) Use the definition of conditional probability to expand:

$P(A_1\cap B)=P(A_1)P(B\mid A_1)=0.30\cdot0.20=0.06$

$P(A_2\cap B)=P(A_2)P(B\mid A_2)=0.55\cdot0.05=0.0275$

(c) By the law of total probability,

$P(B)=P(A_1\cap B)+P(A_2\cap B)=0.06+0.0275=0.0875$

(d) Bayes' theorem says

$P(A_1\mid B)=\dfrac{P(A_1)P(B\mid A_1)}{P(B)}=\dfrac{0.30\cdot0.20}{0.0875}\approx0.686$

$P(A_2\mid B)=\dfrac{P(A_2)P(B\mid A_2)}{P(B)}=\dfrac{0.55\cdot0.05}{0.0875}\approx0.314$

3 0

3 years ago

A factory makes car batteries. The probability that a battery is defective is 1/14. If 400 batteries are tested, about how many

Rufina [12.5K]

Answer:

28.6, that is, about 29 are expected to be defective

Step-by-step explanation:

For each battery, there are only two possible outcomes. Either it is defective, or it is not. The probability of a battery being defective is independent of other betteries. So the binomial probability distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$