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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.

Respuesta :

Answer:

he would make one out of the two shots

Step-by-step explanation:

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:

[tex]P(\frac{B}{A})=\frac{P(B\cap A)}{P(A)}[/tex]

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.

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