Question 582709
A basketball player makes free throws with a 0.7 probability. What is the probability that the player will make at least 4 of the next 5 free throws?
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Binomial with n = 5 and p(makes) = 0.7 and P(misses) = 0.3
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P(x = 4) = 5C4(0.7)^4*0.3 = 5*0.2401*0.03 = 0.3602
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Cheers,
Stan H.
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