Find the probability of getting 2 cherry, 1 lemon
2006-12-10 07:51:17 · 6 answers · asked by staceylynn1913 2 in Education & Reference ➔ Homework Help
There are 11 candies total.
So, first we must find the number of permutations (combinations) of candies we might pull.
For instance, we may pull lemon, lemon, orange...or perhaps...cherry cherry cherry.
To find that, we use the formula:
11! (Where ! is factorial) / (6! x 3! x 2!) since we have 6 cherry, three orange, and 2 lemon.
This gives:
39,916,800 / (720 x 6 x 2)
39,916,800 / 8640 = 4,620
This means there are 4,620 possible combinations of three candies.
We want 2 cherries and 1 lemon, and this could be chosen in the following order.
Cherry / Cherry / Lemon
Lemon / Cherry / Cherry
Cherry / Lemon / Cherry
Clearly, of the 4,620 combinations, only three fit what we want.
The probability of 2 cherry and 1 lemon is:
3/4620 or .0006 or .06% (6 one hundredths of one percent)
Good luck!
Mysstere
2006-12-10 08:05:30 · answer #1 · answered by mysstere 5 · 1⤊ 0⤋
Prob of picking one cherry 6/(6+3+2) = 6/11
Prob of picking second cherry = 5/(5+3+2) = 5/10
Prob fo picking one lemon = 2/(4+3+2) =2/9
The events are mutually exclusive
Prob = 6/11 x 5/10 x 2/9 = 2/33
2006-12-10 08:38:56 · answer #2 · answered by A S 4 · 0⤊ 0⤋
1 out of 15
2006-12-10 07:54:00 · answer #3 · answered by Linda 7 · 0⤊ 0⤋
Total things in bag: 6+3+2 = 11
6 of 11 of those are cherries
2 of 11 of those are lemons
2006-12-10 07:53:52 · answer #4 · answered by Peter the Squirrel 2 · 0⤊ 1⤋
(6+2)/11 * (5+2)/10 * (4+2)/9 = 336/990 =
56/165
2006-12-10 08:07:32 · answer #5 · answered by Barry L 1 · 0⤊ 0⤋
try doing it your self
2006-12-10 07:56:41 · answer #6 · answered by school_girl007 2 · 0⤊ 0⤋