Question 251380: Juanita has 9 rings in her jewelry box. Five are gold and 4 are silver. If she randomly selects 3 rings to wear to a party, find each probabilty.
I have tried added, multiplied variations and don't exactly understand the process.
question: P(exactly 2 silver)
Found 2 solutions by solver91311, edjones:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The probability of getting a silver ring on any given selection is the number of silver rings in the jewelry box at the time the selection is made divided by the total number of rings in the jewelry box at the time the selection is made.
So, there are exactly three equally likely possibilities for the order in which the selection can be made: GSS, SGS, or SSG.
Next, the probability of getting exactly a gold ring then a silver ring and another silver ring is the product of the probability of a gold ring on the first draw,  , times the probability of a silver ring on the second draw,  , times the probability of a silver ring on the third draw,  , times the  probability of this particular order given success at drawing two silver and one gold.
I'll leave it as an exercise for the student to verify that the probability for the other two possible orders of drawing are precisely the same value as the probability for the first order.
The overall probability you seek is the sum (this OR that OR that could happen so + as opposed to this has to happen AND that has to happen so X) of the three probabilities, so simply take the factor out of consideration and the complete answer is:
or, if you prefer:
or, if you prefer:
Whichever way, you get to do your own arithmetic.
John
Answer by edjones(8007)  (Show Source):
You can put this solution on YOUR website! nCr=n!/((n-r)!r!) Combination of n things taken r at a time.
3 gold: 5C3/9C3=10/84
2g, 1s: (5C2 * 4C1)/9C3=10*4 / 84 = 40/84
1g, 2s: (5C1 * 4C2)/9C3= 5*6 /84 = 30/84
3 silver: 4C3/9C3 = 4/84
.
Ed
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