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!)(0.70^{4})(0.30^{3}) = 35(0.006483) = 0.226895

*p*(3) = (7!/3!4!)(0.70^{3})(0.30^{4}) = 35(0.002778) = 0.097241

*p*(2) = (7!/2!5!)(0.70^{2})(0.30^{5}) = 21(0.001191) = 0.025005

_p(_1) = (7!/1!6!)(0.70^{1})(0.30^{6}) = 7(0.000510) = 0.003572

*p*(0) = (7!/0!7!)(0.70^{0})(0.30^{7}) = 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.