Let the probability of success on a Bernoulli trial be 0.20. a. In nine Bernoulli trials,...

Let the probability of success on a Bernoulli trial be 0.20. a. In nine Bernoulli trials, what is the probability that there will be 8 failures? (Round your final answers to 4 decimal places.) Probability b. In nine Bernoulli trials, what is the probability that there will be more than the expected number of failures? (Round your final answers to 4 decimal places.) Probability

Expert Solution

p=probability of success in a Bernoulli trial=0.2

q=1-p=probability of success in a Bernoulli trial=1-0.2=0.8

Let X= number of failures in 9 Bernoulli Trials

1)

The probability that there will be 8 failures in 9 Bernoulli trials =P(X=8)

2)

Expected number of failures= E(X)= n* q= 9*0.8=7.2 ( formula for the expectation of Binomial distribution)

Hence, the probability that there will be more than the expected number of failures in 9 Bernoulli trials=P(X>7.2)=P(X>=8)

( Because X can take only integer values, Being a discrete distribution)

milcah answered 2 years ago

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