Question

A coin jar contains 10 quarters, 6 dimes, and 8 nickels. Find the probability that two quarters are chosen at random from the jar.

Answer 1

Probability that two quarters are chosen at random from the jar is 0.163.

Step-by-step explanation:

We are given that a coin jar contains 10 quarters, 6 dimes, and 8 nickels.

Firstly, Probability of any event =  $\frac{\text{Favorable number of outcomes}}{\text{Total number of outcomes}}$

Also, Probability is never negative and does not exceed 1 which means the value of probability of any event lies between 0 and 1.

Number of ways of selecting 2 quarters from total of 10 quarters is given by = $^{10}C_2$  =  $\frac{10!}{2! \times 8!}$                         {As  $^{n}C_r= \frac{n!}{r! \times (n-r)!}$ }

           =  $\frac{10\times 9\times 8!}{2 \times 8!}$ = 45

Total number of ways of selecting two quarters from the total of 24 items in the jar ( 10 quarters, 6 dimes, and 8 nickels = 24) =  $^{24}C_2$

                           =  $\frac{24!}{2!\times 22!}$  =  276

So, the probability that two quarters are chosen at random from the jar is given by =  $\frac{^{10}C_2 }{^{24}C_2}$  =  $\frac{45}{276}$

                            =  0.163

Hence, the required probability is 0.163.