stephny wrote:hi guys,
please kindly assist.
a jar contains only red, yellow, and orange marbles. if there are 3 red, 5 yellow, and 4 orange marbles, and 3 are chosen from the jar at random without replacing any of them, what is the probability that 2 yellow, 1 red, and no orange marbles will be chosen?
P(exactly n times) = P(one way) * total possible ways.
P(one way):
ONE WAY to select 2 yellow marbles and 1 red marble is YYR.
P(1st marble is Y) = 5/12. (Of the 12 marbles, 3 are Y.)
P(2nd marble is Y) = 4/11. (Of the 11 remaining marbles, 2 are Y.)
P(3rd marble is R) = 3/10. (Of the 10 remaining marbles, 3 are R.)
Since we want all of these events to happen, we MULTIPLY the probabilities:
5/12 * 4/11 * 3/10 = 1/22.
Total possible ways:
YYR is only ONE WAY to get exactly 2 Y's and 1 R.
Now we must account for ALL OF THE WAYS to get exactly 2 Y's and 1 R.
Any arrangement of the letters YYR represents one way to get exactly 2 Y's and 1 R..
Thus, to account for ALL OF THE WAYS to get exactly 2 Y's and 1 R, the result above must be multiplied by the number of ways to arrange the letters YYR.
Number of ways to arrange 3 elements = 3!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical Y's:
3!/2! = 3.
Multiplying the results above, we get:
P(exactly 2 Y's and 1 R) = (1/22) * 3 = 3/22.
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