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A basketball player makes 60% of his free throws. We set him on the line of free-throw and informed him to shoot free throws until he misses. Let the random variable X be the number of free throws taken by the player until he misses. Assuming that his shots are independent, find the probability that he will miss the shot on his 6th throw. Show work detail please
a) 0.04666
b) 0.03110
c) 0.01866
d) 0.00614

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MrRoyal

Answer:

B. 0.03110

Step-by-step explanation:

Given

Probability of Hit = 60%

Required

Determine the probability that he misses at 6th throw

Represent Probability of Hit with P

[tex]P = 60\%[/tex]

Convert to decimal

[tex]P = 0,6[/tex]

Next; Determine the Probability of Miss (q)

Opposite probabilities add up to 1;

So,

[tex]p + q = 1[/tex]

[tex]q = 1 - p[/tex]

Substitute 0.6 for p

[tex]q = 1 - 0.6[/tex]

[tex]q = 0.4[/tex]

Next,is to determine the required probability;

Since, he's expected to miss the 6th throw, the probability is:

[tex]Probability = p^5 * q[/tex]

[tex]Probability = 0.6^5 * 0.4[/tex]

[tex]Probability = 0.031104[/tex]

Hence;

Option B answers the question

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