Find the missing probability. P(A)=15,P(A∪B)=1225,P(A∩B)=7100 ,P(B)=?

Mathematics

1 answer:

Llana [10]2 years ago

6 0

Answer:

Step-by-step explanation:

We will use the addition rule of probability of two events to solve the question. According to the rule given two events A and B;

p(A∪B) = p(A)+p(B) - p(A∩B) where;

A∪B is the union of the two sets A and B

A∩B is the intersection between two sets A and B

Given parameters

P(A)=15

P(A∪B)=1225

P(A∩B)=7100

Required

Probability of event B i.e P(B)

Using the expression above to calculate p(B), we will have;

p(A∪B) = p(A)+p(B) - p(A∩B)

1225 = 15+p(B)-7100

p(B) = 1225-15+7100

p(B) = 8310

Hence the missing probability p(B) is 8310.

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Consider a set of 7500 scores on a national test whose score is known to be distributed normally with a mean of 510 and a standa

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$\mathbb P(X>600)=\mathbb P\left(\dfrac{X-510}{85}>\dfrac{600-510}{85}\right)=\mathbb P(Z>1.059)\approx0.145$

So approximately 14.5% of the scores are higher than 600. This means in a sample of 7500, one could expect to see $0.145\times7500\approx10.86$ scores above 600.