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A basketball player who shoots 80% from the free throw line attempts 2 free throws. Let Event A be a made first attempt and Event B be a made second attempt.

Which statement about the conditional probability is true?



1 The conditional probability of Event B given Event A is P(B|A)=P(B) when two events are not independent.

2 The conditional probability of Event B given Event A is P(B|A)=P(B)P(A) when two events are independent.

3 The conditional probability of Event B given Event A is P(B|A)=P(A)P(B) when two events are independent.

4 The conditional probability of Event B given Event A is P(B|A)=P(A and B)P(A) when two events are not independent.

2 Answers

3 votes

Answer:

he would make one out of the two shots

Explanation:

User Smauel
by
5.3k points
2 votes

Answer with explanation:

Probability of getting shoot from free throw line= 80% =0.80

Probability of not getting shoot from free throw line = 100% - 80% = 20% = 0.20

A student attempt throwing ,twice from free throw line.

A= First Attempt

B= Second Attempt

If Events , A and B are Independent,then Probability of A and B is given as

P (A ∩ B)= P(A) × P(B)

→And,Conditional probability of event A has definitely occurred and then probability that event B will occur,when these two events A and B are not Independent, is given by:


P((B)/(A))=(P(B\cap A))/(P(A))

Option 4: The conditional probability of Event B given Event A is P(B|A)=P(A and B)/P(A) when two events are not independent.

User BluePearl
by
5.6k points