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A college conducts a survey and finds that 45% of students call home every weekend. Ten random students are selected. 1) What is the probability that exactly 5 students will say they call home every weekend. 2) What is the probability that all the students will say they call home every weekend

a) P(exactly 5) = 0.207, P(all) = 0.005
b) P(exactly 5) = 0.200, P(all) = 0.002
c) P(exactly 5) = 0.225, P(all) = 0.001
d) P(exactly 5) = 0.210, P(all) = 0.003

User Robor
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1 Answer

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Final answer:

To answer the probability questions, one must apply the binomial probability formula. The probability that exactly 5 out of 10 students say they call home is approximately 0.201, and the probability that all 10 students say they call home is approximately 0.000. Option b) is the closest to the computed values with P(exactly 5) = 0.200 and P(all) = 0.002.

Step-by-step explanation:

The question pertains to probability and binomial distribution, which is a part of Mathematics. Specifically, this falls under the concepts typically covered in a college-level statistics course.

To find the probability that exactly 5 students will say they call home every weekend, one can use the binomial probability formula:

P(X = k) = nCk * p^k * (1-p)^(n-k)

Where:

  • n is the number of trials (students), which is 10,
  • k is the number of successes (students who say they call home), which is 5,
  • p is the probability of success on a single trial (45% or 0.45 in this case), and
  • (1-p) is the probability of failure on a single trial.

To calculate this:

nCk = 10! / [5! * (10-5)!] = 252, and then

P(X = 5) = 252 * (0.45)^5 * (0.55)^5 = 0.200658124

The probability that all the students say they call home every weekend is determined by p^n because all ten students (n) must be successes (p), thus:

P(X = 10) = 0.45^10 = 0.0003405063

Rounding to three decimal places:

P(exactly 5) = 0.201, P(all) = 0.000

Therefore, the closest answer to the provided options would be:

b) P(exactly 5) = 0.200, P(all) = 0.002

User Dave Taylor
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