Suppose there are 8 green tags, 12 white tags and 4 red tags left to use as name tags at a conference...

Question

Suppose there are 8 green tags, 12 white tags and 4 red tags left to use as name tags at a conference. If tags are given out randomly at the registration desk,

a) what is the probability that the first two tags given out are red?

b) what is the probability that if part a did happen, then more than 3 out of the next 5 tags given out would be white?

Answer 1

there are 4 ways to get red on the first time out of 24 possibilities 4/(8+12+4) = 4/24 = 1/6

on the 2nd time, there are 3 ways out of 23 or 3/23. Multiply the two probabilities: 1/6 x 3/23 = 1/46=about 2.17% chance of 2 reds on first two

IF you did get 2 reds on the first two tries, odds of getting more than 3 whites on the next 5 tries is 29/133 = about 21.8%

there's 4/133 ways to get all 5 whites: 12/22 x 11/21 x 10/20 x 9/19 x 8/18

there's 25/133 ways to get 4 out 5 whites on the next 5 tries or 5 times 12/22 x 11/21 x 10/20 x 9/19 x 10/18

(4+25)/133 = 29/133 = about 21.8%