Involving computation of probabilities for Bernoulli trials. (1) At least 4 succ

Question

Involving computation of probabilities for Bernoulli trials.

(1) At least 4 successes in 10 trials with p = 0.8 / My wrong answer = 0.998619648

(2) At least 4 failures in 6 trials with p = 0.3

Explanation / Answer

Dear Student Thank you for using Chegg !! Given a) P (success) = 0.8 P (Failure) = 0.2 Now there have been 10 number of trials and we have to find the probability of atleast 4 success => There will be cases with 4 success, 5 success , 6 success upto 10 success and in case 4 success there will be 6 failure and subsequently for 5 success 5 failure and so on P (4 ) (0.8^4)*(0.2^6) = 2.62144E-05 P(5 ) (0.8^5)*(0.2)^5) = 0.000104858 P(6) (0.8^6)*(0.2)^4) = 0.00041943 P(7) (0.8^7)*(0.2)^3) = 0.001677722 P(8) (0.8^8)*(0.2)^2) = 0.006710886 P(9) (0.8^9)*(0.2)^1) = 0.026843546 P(10) (0.8^10)*(0.2)^0) = 0.107374182 Final probability = Sum of all the cases = 0.143156838 Solution

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