SOLUTION: A box contains 21 yellow, 24 green and 27 red jelly beans.
If 13 jelly beans are selected at random, what is the probability that:
a) 10 are yellow?
b) 10 are yellow and
Algebra ->
Probability-and-statistics
-> SOLUTION: A box contains 21 yellow, 24 green and 27 red jelly beans.
If 13 jelly beans are selected at random, what is the probability that:
a) 10 are yellow?
b) 10 are yellow and
Log On
Question 710221: A box contains 21 yellow, 24 green and 27 red jelly beans.
If 13 jelly beans are selected at random, what is the probability that:
a) 10 are yellow?
b) 10 are yellow and 2 are green?
c) At least one is yellow?
Sorry but i dont have the slightest clue how to do this.I would aprreciate the help! Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A box contains 21 yellow, 24 green and 27 red jelly beans.
There are 72 jelly beans.
----------------------------
If 13 jelly beans are selected at random, what is the probability that:
a) 10 are yellow?
# of ways to succeed: 21C10*51C3
# of possible outcomes: 72C13
---
P(10 yellow) = [21C10*51C3]/72C13 = 0.0001036...
=======================================================
b) 10 are yellow and 2 are green?
P(10 yellow and 2 green) = [21C10*24C2*29C1]/72C13 = 0.000039814
------------------------------------------
c) At least one is yellow = 1 - P(none are yellow)
= 1 - [51C13/72C13]
---
= 0.9933
==============
Cheers,
Stan H.
===============
(