How can I find the probability (in fractions)? S={3,5,6,8,9,12,13,14,15,16} A={3,12,5,13} B={14,15,6,3} C={12,5,8,9,16} 1. P(B|A)= ? and are...

I am assuming that we are selecting one item from a sample space of

s={3,5,6,8,9,12,13,14,15,16}.

We have event spaces:

A={3,5,12,13}

B={3,6,14,15}

C={5,8,9,12,16}

We can compute the probabilities for each event: P(A)=P(B)=2/5 and P(C)=1/2. (We find these by taking the size of the event space divided by the size of the sample space.)

(1) P(B|A) is the probability that the item selected is in B given that it is in A. If the item selected is...

I am assuming that we are selecting one item from a sample space of

s={3,5,6,8,9,12,13,14,15,16}.

We have event spaces:

A={3,5,12,13}

B={3,6,14,15}

C={5,8,9,12,16}

We can compute the probabilities for each event: P(A)=P(B)=2/5 and P(C)=1/2. (We find these by taking the size of the event space divided by the size of the sample space.)

(1) P(B|A) is the probability that the item selected is in B given that it is in A. If the item selected is in A then it is one of 3,5,12,13. Only 3 is in B, so the probability that an item is in B given that it is in A is 1/4.

P(B|A)=1/4. Using the formula `P(B|A)=(P(A and B))/(P(A)) ` , and noting that the probability that an item is in A and in B is 1/10 (only 3 is in both A and B, so there is 1 item out of the 10 total items.)

So P(B|A)=(1/10)/(2/5)=1/4 as above.

The events are not independent since `P(A and B) != P(A)*P(B) `

(2) P(A or B) These events are not mutually exclusive (it is possible for an item to be in A and B) so P(A or B)=P(A)+P(B)-P(A and B) or

P(A or B)=2/5+2/5-1/10=7/10

(3) P(A|C)

`P(A|C)=(P(A and C))/(P(C))=(1/5)/(1/2)=2/5 `

(If the item is known to be in C then it is one of 5,8,9,12,16; two of these numbers are in A so the probability is 2/5 as above.)