Question 1120948
this problem can be viewed as a binomial experiment, since there are two possible outcomes for the patient and the probability (0.40) stays constant
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binomial probability formula is
:
Probability (P) (k successes in n trials) = nCk * p^k * (1-p)^(n-k), where nCk = n! / (k! * (n-k)!)
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p = 0.40
:
(i) P (1 success in 6 trials) = 6C1 * (0.40)^1 * (1-0.40)^(6-1) = 0.1866
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(ii) P (k < or = 2 in 6 trials) = 0.5443
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(iii) P (k > or = 2 in 6 trials) = 0.7667
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(iv) P (k = 3 or 4 in 6 trials) = P(k = 3 in 6 trials) + P(k = 4 in 6 trials) = 0.2764 + 0.1382 = 0.4146
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