a basketball player makes 60% of the free throws that he attempts, except that if he has just tried and missed a free throw then his chances of making a second one go down to only 30%. what is the probability that he makes both attempts? what is the probability that he makes at least one attempt?
To make both: .6 times .6 = .36
Make at least one of two:
1st 2nd
make .6 make .6 miss .4
miss .4 make .3 miss .7
Makes at least one means one or two made
(.6)(.6) + (.6)(.4) + (.4)(.3
multiply and sum and you should get your answer.
To find the probabilities, let's break down the situation step by step.
Probability of making both attempts:
First, we need to determine the probability that the player makes the first free throw (FT1) and the second free throw (FT2).
1. Probability of making FT1:
Since the player has a 60% chance of making a free throw, the probability of making FT1 is 0.6.
2. Probability of making FT2:
If the player made FT1, then for the second attempt his chances of making it drop to 30%. So, the probability of making FT2 after making FT1 is 0.3.
To find the probability of both events happening (making FT1 and then making FT2), we multiply the probabilities:
P(make FT1 and make FT2) = P(make FT1) * P(make FT2|make FT1)
= 0.6 * 0.3
= 0.18
Therefore, the probability that the player makes both attempts is 0.18 or 18%.
Probability of making at least one attempt:
To find the probability of making at least one attempt, we need to consider two scenarios: either the player makes both attempts OR the player misses the first attempt but makes the second one.
1. Probability of making both attempts:
We have already found this probability to be 0.18 or 18%.
2. Probability of missing the first attempt and making the second attempt:
The probability of missing the first attempt is (1 - the probability of making the first attempt), which is 1 - 0.6 = 0.4. After missing the first attempt, the probability of making the second attempt is 0.3.
To find the probability of this scenario, we multiply the probabilities:
P(miss FT1 and make FT2) = P(miss FT1) * P(make FT2|miss FT1)
= 0.4 * 0.3
= 0.12
Therefore, the probability of making at least one attempt is the sum of the probabilities that the player makes both attempts and the player misses the first attempt and makes the second one:
P(make at least one attempt) = P(make FT1 and make FT2) + P(miss FT1 and make FT2)
= 0.18 + 0.12
= 0.3
Therefore, the probability that the player makes at least one attempt is 0.3 or 30%.
To calculate the probability that the basketball player makes both attempts, we can use the following formula for the probability of two independent events occurring:
P(A and B) = P(A) * P(B)
The probability that the player makes the first free throw is 60% or 0.60. The probability that the player makes the second free throw, given that he made the first one, is also 60% or 0.60.
P(makes both attempts) = P(makes first attempt) * P(makes second attempt)
= 0.60 * 0.60
= 0.36 or 36%
Therefore, the probability that the basketball player makes both attempts is 36%.
To calculate the probability that the basketball player makes at least one attempt, we can use the complementary probability approach. It means we calculate the probability of the event not happening and subtract it from 1.
The player can either make both attempts or miss at least one. So,
P(makes at least one attempt) = 1 - P(misses both attempts)
The probability that the player misses the first free throw is 1 - 0.60 = 0.40. The probability that the player misses the second free throw, given that he missed the first one, is 1 - 0.30 = 0.70.
P(misses both attempts) = P(misses first attempt) * P(misses second attempt)
= 0.40 * 0.70
= 0.28 or 28%
Therefore, the probability that the basketball player makes at least one attempt is:
P(makes at least one attempt) = 1 - P(misses both attempts)
= 1 - 0.28
= 0.72 or 72%