# Counting and Probability Help

• June 18th 2009, 06:34 PM
ajmaal14
Counting and Probability Help
okay so I have a test tomorrow and I was doing the review questions. I am stock on 3 of the questions. It be greatly appreciated if you guys can help. Here are the questions:

1. in a bag of candy canes, there are 14 broken candies, mixed in with 20 good ones. if neha chooses 3 candies at random from the bag what is the probability that all 3 are good

2. Think of a word in the English language in which the probability of selecting a vowel is 3/5.

3. In how many ways can a group of five people be chosen from seven couples (each of which has one more and one female) to form a club, given each of the following conditions?
A) All are equally eligible for the club
B) The club must include two females and 3 males.
Thanks
• June 18th 2009, 06:46 PM
apcalculus
Quote:

Originally Posted by ajmaal14
okay so I have a test tomorrow and I was doing the review questions. I am stock on 3 of the questions. It be greatly appreciated if you guys can help. Here are the questions:

1. in a bag of candy canes, there are 14 broken candies, mixed in with 20 good ones. if neha chooses 3 candies at random from the bag what is the probability that all 3 are good

2. Think of a word in the English language in which the probability of selecting a vowel is 3/5.

3. In how many ways can a group of five people be chosen from seven couples (each of which has one more and one female) to form a club, given each of the following conditions?
A) All are equally eligible for the club
B) The club must include two females and 3 males.
Thanks

Some help with the first two:

1)
P(first one is good) = 20/34
P(second one is good) = 19/33
P(third one is good) = 18/32

Multiply all three probabilities since the events are independent.

2) route
above

Good luck!!
• June 18th 2009, 06:51 PM
putnam120
1)
There are a total of 34 candies in the bag. The probability that the first is good is 20/34. Thus leaving 33, so the probability the second is good is now 19/33. By similar logic the probability for the third being good is 18/32. Thus the final answer is $\frac{20\cdot 19\cdot 18}{34\cdot 33\cdot 32}$
• June 18th 2009, 06:52 PM
ajmaal14
Thanks a lot just dont get the multiplying part
Thanks a lot man
k soo do 20/34 x 19/33 x 18/32
if i do that i get 285/1496