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anonymous
 5 years ago
What is the probability of obtaining exactly 4 successes in 5 trials when the probability of success is 0.5.
Round your answer to 4 decimal places.
anonymous
 5 years ago
What is the probability of obtaining exactly 4 successes in 5 trials when the probability of success is 0.5. Round your answer to 4 decimal places.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm not so sure but i think the probability of success is 0.5, then the probability of not success is 10.5=0.5 obtaining 4 successes in 5 trials = 4 successes and 1 not success so, the probability of it is 0.5^4 x 0.5 = 0.5^5 = 1/32

GG
 5 years ago
Best ResponseYou've already chosen the best response.0I think you are right. Those happenings are independent, so we just multiply probabilities of each.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats actually not the answer its (5)(0.5)^4(0.5)^54=0.1563

GG
 5 years ago
Best ResponseYou've already chosen the best response.05 times (0.5^4) times (0.5^5) 4 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm maybe because there are 5 trials, the 1/32 that i got before should be multiplied by 5, therefore the answer is the same as yours ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/32 x 5 = 0.03125 x 5 = 0.15625

GG
 5 years ago
Best ResponseYou've already chosen the best response.0but why to multiply it by 5?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm also confused, but i think it's because there are 5 trials

GG
 5 years ago
Best ResponseYou've already chosen the best response.0but if we have 100 trials, than the possibility would be 3.1, and that is impossible, right? so, that is not the reason :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i figured it out by the book saying this Let T denote the outcome “Tail shows” (success) and let H denote the outcome “Head shows” (failure). Using Formula (2), in which k = 1 (one success), n = 6 (the number of trials), and (the probability of success using a fair coin), we obtain

GG
 5 years ago
Best ResponseYou've already chosen the best response.0P= (n!/[k!(nk)!] ) p^k * q^(nk) use this formula. n is number of trials, k is number of successes, p is possibility of success, q=1p :) is it ok now? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yes, combination. 5C4 0.5^4 * 0.5. thank you
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