Question

Joaquin has 6 dimes and 4 quarters in his pocket. If he randomly selects 4 coins (simultaneously), find the probability that he will choose exactly 1 dime or exactly 1 quarter?

Answer 1

Using the hypergeometric distribution, it is found that there is a 0.4953 = 49.53% probability that he will choose exactly 1 dime or exactly 1 quarter.

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

Hypergeometric distribution:  

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

$C_{n,x} = \frac{n!}{x!(n-x)!}$  

The parameters are:

• 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.  

In this problem:

• 6 + 4 = 10 coins, thus $N = 10$.
• 4 are chosen, thus $n = 4$.
• 6 are dimes, thus $k = 6$.
• The probability of exactly 1 dime is P(X = 1).
• 1 quarter is 4 - 1 = 3 dimes, thus the probability of exactly one quarter is P(X = 3).

Then:

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

$P(X = 1) = h(1,10,4,6) = \frac{C_{6,1}C_{4,3}}{C_{10,4}} = 0.1143$

$P(X = 3) = h(3,10,4,6) = \frac{C_{6,3}C_{4,1}}{C_{10,4}} = 0.381$

$p = P(X = 1) + P(X = 3) = 0.1143 + 0.381 = 0.4953$

0.4953 = 49.53% probability that he will choose exactly 1 dime or exactly 1 quarter.

A similar problem is given at https://brainly.com/question/4818951