Question

Suppose you have 4 UCR shirts and 7 shirts from other universities that you got when trying to decide where to go to college. Whilst packing for a weekend of camping with friends, you you reach into your drawer full of university shirts and pull three out at random. What is the probability that all three are UCR shirts?

Answer 1

Answer:

0.0242 = 2.42% probability that all three are UCR shirts.

Step-by-step explanation:

The shirts are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

In this question:

4 + 7 = 11 shirts mean that $N = 11$

4 UCR shirts mean that $k = 4$

3 are pulled, which means that $n = 3$

What is the probability that all three are UCR shirts?

This is $P(X = 3)$. So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,11,3,4) = \frac{C_{4,3}*C_{7,0}}{C_{11,3}} = 0.0242$

0.0242 = 2.42% probability that all three are UCR shirts.

Answer 2

The probability that all three are UCR shirts is 0.0242 or 2.42%.

Given that,

Suppose you have 4 UCR shirts and 7 shirts from other universities that you got when trying to decide where to go to college.

Whilst packing for a weekend of camping with friends, you reach into your drawer full of university shirts and pull three out at random.

We have to determine,

What is the probability that all three are UCR shirts.

According to the question,

The probability of x successes is given by the following formula:

$P (X=x) = \ (h,k,x,N) = \ \dfrac{^kC_x \ \times \ ^{N-k} C_n_-_k} {^NC_n}$

Where, x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Then,

N = 4 + 7 = 11 shirts  , k =4 UCR shirts , n = 3 are pulled random

Therefore,

Substitute the values in the formula,

$P (X=x) = \ (n,k,x,N) = \ \dfrac{^kC_x \ \times \ ^{N-k} C_n_-_k} {^NC_n}\\\\P (X=x) = \ (3,4,3,11) = \ \dfrac{^4C_3 \ \times \ ^{11-4} C_3_-_k} {^11C_3}\\\\P (X=3) = \ (3,4,3,11) = \ \dfrac{^4C_3 \ \times \ ^{7} C__0} {^{11}C_3}\\\\P(X=3) = P (X=3) = \ (3,4,3,11) = \dfrac{4}{1320}\\\\P (X=3) = \ (3,4,3,11) = 0.0242$

Hence, The probability that all three are UCR shirts is 0.0242 or 2.42%.

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