In Lloyd's pocket are 6 coins: 3 dimes, 2 nickels, and 1 penny. If Lloyd randomly takes two coins out of his pocket, what is the probablity that he will take out 11 cents? The book shows the answer 1/5, but my question is wouldn't the number of each coin come into play?
c(3,1)c(1,1)/c(6,2) use cpr fn. in calculator
To calculate the probability of Lloyd taking out 11 cents, we need to consider the total number of possible outcomes and the number of favorable outcomes.
Total Number of Possible Outcomes:
Lloyd has 6 coins in total. When he takes out two coins, he has 6 choices for the first coin and 5 choices for the second coin. Therefore, there are a total of 6 x 5 = 30 possible outcomes.
Number of Favorable Outcomes:
Lloyd will have a total of 11 cents if he takes out a dime and a penny, or a nickel and a nickel.
For the scenario of taking out a dime and a penny:
Lloyd has 3 dimes to choose from, and after taking one dime out, he will have 2 pennies left in the pocket. Therefore, there are 3 x 2 = 6 favorable outcomes for this scenario.
For the scenario of taking out a nickel and a nickel:
Lloyd has 2 nickels to choose from, and after taking one nickel out, he will have 1 nickel left in the pocket. Therefore, there are 2 x 1 = 2 favorable outcomes for this scenario.
Total Number of Favorable Outcomes:
Adding the favorable outcomes from both scenarios, we have a total of 6 + 2 = 8 favorable outcomes.
Probability:
The probability of an event is given by the ratio of favorable outcomes to total outcomes. Therefore, the probability of Lloyd taking out 11 cents is 8/30.
Simplifying the fraction, we have:
8/30 = 4/15
So, the probability of Lloyd taking out 11 cents is 4/15, not 1/5 as mentioned in the book.
The book seems to have made an error, possibly by overlooking the number of each coin in Lloyd's pocket. The correct answer takes into account the number of each coin and gives a probability of 4/15.