# Probability Distributions - Quicker method?

Is there a quicker method to getting the correct answer? The questions is easy but the time it takes to get the answer is long. (Took me 2:47 and that was using the STO/RCL functions on TI BA II Plus) Just curious. Thanks

P = .7 , N = 7, X = 4 Find probability of 4 or fewer successes.

The correct answer is 35.3 percent.

The observed success rate is 4/7 = 0.571, or 57.1 percent. The probability of four or fewer successes is F(4) = p(4) + p(3) + p(2) + p(1) + p(0), where p(4), p(3), p(2), p(1), and p(0) are respectively the probabilities of 4, 3, 2, 1, and 0 successes, according to the binomial distribution with n = 7 and p = 0.70. We have

p(4) = (7!/4!3!)(0.704)(0.303) = 35(0.006483) = 0.226895

p(3) = (7!/3!4!)(0.703)(0.304) = 35(0.002778) = 0.097241

p(2) = (7!/2!5!)(0.702)(0.305) = 21(0.001191) = 0.025005

_p(_1) = (7!/1!6!)(0.701)(0.306) = 7(0.000510) = 0.003572

p(0) = (7!/0!7!)(0.700)(0.307) = 1(0.000219) = 0.000219

Summing all these probabilities, you conclude that F(4) = 0.226895 + 0.097241 + 0.025005 + 0.003572 + 0.000219 = 0.352931, or 35.3 percent.

1- pr(5 or more successes).