Notice that since after one ball is selected it is not replaced by another one, the events are dependent. Recall how we calculate the probability of two dependent events.
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Practice makes perfect
The ball bin in the gym contains 8 soccer balls, 12 basketballs, and 10 kickballs. Coach Meyers selects 2 balls at random without replacing the first. We want to find the probability that 2 basketballs were selected. Since the first ball is not replaced, the events are dependent. Recall how we calculate the probability of two dependent events.
P(AandB)=P(A) * P(B|A)
Let's mark the event of selecting a basketball as B. The first and the second event in this case are identical, since we want 2 basketballs to be chosen. We are interested in the following probability.
P(BandB)=P(B) * P(B|B)
Let's start with calculating P(B). To do so, we need to divide the number of favorable outcomes by the number of all possible outcomes. There are 12 basketballs, and when the coach chooses the first ball there are 8+12+10= 30 balls in the ball bin. Therefore, our probability will be the quotient of these numbers.
Now let's move on to the second probability, P(B|B). This is a conditional probability stating that a basketball was chosen as second given that a basketball was also chosen first. In this case we have 12-1= 11 basketballs, since one was already chosen. Also as a result, this time we will choose from 30-1= 29 balls.
P(B|B)=11/29
Finally we can multiply the two obtained probabilities to find the probability of events B and B.
