Question

The probability that an Oxnard University student is carrying a backpack is .70. If 10 students are observed at random, what is the probability that fewer than 7 will be carrying backpacks

Answer 1

Answer:

35.03% probability that fewer than 7 will be carrying backpacks

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they are carrying a backpack, or they are not. The probability of a student carrying a backpack is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that an Oxnard University student is carrying a backpack is .70.

This means that $p = 0.7$

If 10 students are observed at random, what is the probability that fewer than 7 will be carrying backpacks

This is $P(X < 7)$ when $n = 10$. So

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.7)^{0}.(0.3)^{10} = 0.000006$

$P(X = 1) = C_{10,1}.(0.7)^{1}.(0.3)^{9} = 0.0001$

$P(X = 2) = C_{10,2}.(0.7)^{2}.(0.3)^{8} = 0.0014$

$P(X = 3) = C_{10,3}.(0.7)^{3}.(0.3)^{7} = 0.0090$

$P(X = 4) = C_{10,4}.(0.7)^{4}.(0.3)^{6} = 0.0368$

$P(X = 5) = C_{10,5}.(0.7)^{5}.(0.3)^{5} = 0.1029$

$P(X = 6) = C_{10,6}.(0.7)^{6}.(0.3)^{4} = 0.2001$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.000006 + 0.0001 + 0.0014 + 0.0090 + 0.0368 + 0.1029 + 0.2001 = 0.3503$

35.03% probability that fewer than 7 will be carrying backpacks