The probability that a randomly chosen sales prospect will make a purchase is 20%. What is the probability (to three decimal places) that the salesperson will make four or more sales if six sales calls are made on a given day

1.7%

Step-by-step explanation:

We have to calculate the probability that the salesperson will make four or more sales if six sales calls are made on a given day, that is:

P (x => 4)

Therefore, we must calculate when x = 4, when x = 5, and when x = 6 and add. p = 0.2, n = 6

P (x = r) = nCr * p ^ r * (1 - p) ^ (n-r)

Also, nCr = n! / (r! * (n-r) !, now replacing:

P (x = 4) = 6! / (4! * (6-4)! * 0.20 ^ 4 * 0.80 ^ (6-4)

P (x = 4) = 15 * 0.001024 = 0.01536

P (x = 5) = 6! / (5! * (6-5)! * 0.20 ^ 5 * 0.80 ^ (6-5)

P (x = 5) = 6 * 0.000256 = 0.001536

P (x = 6) = 6! / (6! * (6-6)! * 0.20 ^ 6 * 0.80 ^ (6-6)

P (x = 6) = 1 * 0.000064 = 0.000064

now,

P (x => 4)  = P (x = 4) + P (x = 5) + P (x = 6)

P (x => 4)  = 0.01536) + 0.001536 + 0.000064

P (x => 4) = 0.01696 = 0.017

It means that the probability is 1.7%

y=sec x is the correct answer

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