Question

Let S be a sample space and E and F be events associated with S. Suppose that

Pr left parenthesis Upper E right parenthesis equals 0.6Pr(E)=0.6​,

Pr left parenthesis Upper F right parenthesis equals 0.2Pr(F)=0.2

and

Pr left parenthesis Upper E intersect Upper F right parenthesis equals 0.1Pr(E∩F)=0.1.

Calculate the following probabilities.

​(a) Pr left parenthesis E|F right parenthesisPr(E|F)

​(b) Pr left parenthesis F|E right parenthesisPr(F|E)

​(c) Pr left parenthesis E| Upper F prime right parenthesisPrE|F′

​(d) Pr left parenthesis Upper E prime | Upper F prime right parenthesisPrE′|F′

Answer 1

Answer:

a) P(E|F) = 0.5

b) P(F|E) = 0.167

c) P(E|F') = 0.625

d) P(E′|F′) = 0.375

Step-by-step explanation:

P(E) = 0.6

P(F) = 0.2

P(E n F) = 0.1

a) P(E|F) = Probability of E occurring, given F has already occurred. It is given mathematically as

P(E|F) = [P(E n F)]/P(F) = 0.1/0.2 = 0.5

b) P(F|E) = Probability of F occurring, given E has already occurred. It is given mathematically as

P(F|E) = [P(E n F)]/P(E) = 0.1/0.6 = 0.167

c) P(E|F′) = Probability of E occurring, given F did not occur. It is given mathematically as

P(E|F') = [P(E n F')]/P(F')

But P(F') = 1 - P(F) = 1 - 0.2 = 0.8

P(E n F') = P(E) - P(E n F) = 0.6 - 0.1 = 0.5

P(E|F') = 0.5/0.8 = 0.625

d) P(E′|F′) = [P(E' n F')]/P(F')

P(F') = 0.8, P(Universal set) = P(U) = 1

P(E' n F') = P(U) - [P(E n F') + P(E' n F) + P(E n F) = 1 - (0.5 + 0.1 + 0.1) = 0.3

P(E′|F′) = 0.3/0.8 = 0.375

Answer 2

(a) The probability of occurring of E, with F already occurred P(E|F) is 0.5.

(b) The Probability of occurring of F, given E has already occurred is 0.167.

(c)  The probability of occurring of E, given F did not occur is 0.625.

(d)  The required value of P(E′|F′) is 0.375.

Given data:

The sample space is, S.

The events are E and F.

And, probability of happening the event E is, P(E) = 0.6.

The probability of happening the event F is, P(F) = 0.2.

The probability of E intersection F is, $P(E \cap F) = 0.1$

(a)

P(E|F) means that Probability of E occurring, given F has already occurred. It is given mathematically as,

$P(E|F)=\dfrac{P (E\cap F}{P(F)}$

Solving as,

$P(E|F)=\dfrac{0.1}{0.2}\\\\P(E|F)=0.5$

Thus, the probability of occurring of E, with F already occurred P(E|F) is 0.5.

(b)

P(F|E) means the Probability of occurring of F, given E has already occurred. It is given mathematically as,

$P(F|E)=\dfrac{P (E\cap F)}{P(E)}\\\\P(F|E)=\dfrac{0.1}{0.6}\\\\P(F|E)=0.167$

Thus, the Probability of occurring of F, given E has already occurred is 0.167.

(c)

P(E|F′) means that  probability of occurring of E, given F did not occur. It is given mathematically as

$P(E|F')=\dfrac{P (E\cap F')}{P(F')}$  

And,

P(F') = 1 - P(F) = 1 - 0.2 = 0.8

P(E n F') = P(E) - P(E n F)

P(E n F') = 0.6 - 0.1 = 0.5.

So,

$P(E|F')=\dfrac{0.5}{0.8}\\\\P(E|F')=0.625$

Thus, the probability of occurring of E, given F did not occur is 0.625.

 

(d)

Now, we know that

P(E′|F′) = [P(E' n F')]/P(F')

Where,

P(F') = 0.8 and  P(U) = 1

Because P(U) denotes the universal set.

So, the values can be calculated as,

P(E' n F') = P(U) - [P(E n F') + P(E' n F) + P(E n F)

P(E' n F') = 1 - (0.5 + 0.1 + 0.1) = 0.3

P(E′|F′) = 0.3/0.8 = 0.375

Thus, the required value of P(E′|F′) is 0.375.

Learn more about the probability distribution here:

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