Question 1133729
Use the binary probability formula
:
Probability(P) (k successes in n trials) = nCk * p^k * (1-p)^(n-k), where nCk = n!/(k! * (n-k)!), p is probability of a success
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p is given to be 0.03
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a) P(k = 2 in 2 trials) = 2C2 * (0.03)^2 * (1-0.03)^(2-2) = 0.0009
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b) P(k = 1 or 2 in 2 trials) = P(k = 1 in 2 trials) +P(k=2 in 2 trials) = 0.0582 +0.0009 = 0.0591
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c) P(k = 10 in 10 trials) = 10C10 * (0.03)^10 * (1-0.03)^(10-10) = 0
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Note a success is a failure for this problem
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