The probability of contracting a stomach virus while visiting Mexico is 60%. Find the probability that among 15 students visiting Mexico, a. More than one student contracts a stomach virus:b. More than 2 students contract a stomach virus ...

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Step-by-step explanation:

You multiply the probability of contracting the virus by the number of students.

Answer: a.) 0.999972

b.) 0.999719

Step-by-step explanation:

For a visiting student to contract the virus, the probability of success, p, is given as 60% = 0.6

This means probability of failure, q, is 40% = 0.4.

Number of sample n is given as 15.

To solve this question, we use the probability distribution Formula for selection. By this, the probability is given as

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

a. WhereP (x > 1) = 1 - P (x < / = 1) i. e when x = 0 and 1

When x=0,

P (x=0) = 15C0 * 0.6^0 * 0.4 ^ 15.

P (x=0) = 1 * 1 * 0.000001 = 0.000001.

When x=1

P (x=1) = 15C1 * 0.6^1 * 0.4^14

P (x=1) = 15 * 0.6 * 0.000003

P (x=1) = 0.000027

Hence, probability of x< / = 1 is given as

P (x=0) + P (x=1) = 0.000001 + 0.000027 = 0.000028

Hence, probability of x>1 = P (x>1) = 1-0.000028

P (x>1) = 0.999972.

Probability that more than one student contracts the virus = 0.999972.

b.) probability that more that 2student contracts the virus = P (x>2)

Like the previous solution, we determine for probability of when x less than or equal to 2 i. e when p (x< / = 2), then subtract our answer from 1.

P (x>2) = 1 - P (x< / = 2)

Since we know P (x=0) and P (x=1) from our previous solution, we just find for the P (x=2)

P (x=2) = 15C2 * 0.6^2 * 0.4^13

P (x=2) = 105 * 0.36 * 0.0000067

P (x=2) = 0.000253.

P (x
P (x
Hence probability that more than 2 students contract the virus, that is:

P (x>2) = 1 - 0.000281

P (x>2) = 0.999719