A jar contains 2 pennies, 3 nickels and 6 dimes. A child selects 2 coins at random without replacement from the jar. Let X represent the amount in cents of the selected coins.
Round your answers to 3 decimal places.
Find the probability X = 10.
Find the probability X = 11.
P(nn) = 3/11 * 2/10
That is the only way to get 10 cents in 2 coins
To get 11 cents, you have
P(pd) = P(dp) = 6/11 * 2/10
so P(11 cents) = 2 * 6/11 * 2/10
To find the probability that the sum of the selected coins is 10 or 11, we need to find the total number of possible outcomes and the number of favorable outcomes.
Total number of possible outcomes:
The child selects 2 coins from a jar that contains a total of 2 + 3 + 6 = 11 coins. So the total number of possible outcomes can be calculated using combinations: C(11, 2) = 55. This means there are 55 different pairs of coins the child can select.
Number of favorable outcomes for X = 10:
To find the number of favorable outcomes for X = 10, we need to consider all the possible pairs of coins that add up to 10. We can have either two pennies or a nickel and five dimes.
1. Two pennies: There are 2 pennies in the jar, so the child can select them in C(2, 2) = 1 way.
2. A nickel and five dimes: There are 3 nickels and 6 dimes in the jar, so the child can select one nickel in C(3, 1) = 3 ways, and five dimes in C(6, 5) = 6 ways. This gives us a total of 3 * 6 = 18 ways.
Therefore, the number of favorable outcomes for X = 10 is 1 + 18 = 19.
Number of favorable outcomes for X = 11:
To find the number of favorable outcomes for X = 11, we need to consider all the possible pairs of coins that add up to 11. We can have either a nickel and six dimes or two nickels and one dime.
1. A nickel and six dimes: There are 3 nickels and 6 dimes in the jar, so the child can select one nickel in C(3, 1) = 3 ways, and six dimes in C(6, 6) = 1 way. This gives us a total of 3 * 1 = 3 ways.
2. Two nickels and one dime: There are 3 nickels and 6 dimes in the jar, so the child can select two nickels in C(3, 2) = 3 ways, and one dime in C(6, 1) = 6 ways. This gives us a total of 3 * 6 = 18 ways.
Therefore, the number of favorable outcomes for X = 11 is 3 + 18 = 21.
Now, we can calculate the probabilities:
Probability X = 10:
P(X = 10) = (Number of favorable outcomes for X = 10) / (Total number of possible outcomes) = 19/55 ≈ 0.345.
Probability X = 11:
P(X = 11) = (Number of favorable outcomes for X = 11) / (Total number of possible outcomes) = 21/55 ≈ 0.382.
So the probability of X = 10 is approximately 0.345, and the probability of X = 11 is approximately 0.382.