kornerua wrote:I need help understanding this:
From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?
Alternate approach:
P(good outcome) = P(one way) * (number of possible ways).
Let K = Kori, J = Jason, and N = a member who is not K or J.
P(one way):
One way to choose K and J is as follows:
KJNN.
P(K is selected first) = 1/7. (Of the 7 members, only one is K.)
P(J is selected second) = 1/6. (Of the 6 remaining members, only one is J.)
P(N is selected third) = 5/5. (Of the 5 remaining members, all 5 are not K or J.)
P(N is selected fourth) = 4/4. (Of the 4 remaining members, all 4 are not K or J.)
To combined these probabilities, we multiply:
1/7 * 1/6 * 5/5 * 4/4 = 1/42.
Total possible ways:
KJNN is only ONE WAY that K and J can be selected.
Now we must account for ALL OF THE WAYS that K and J can be selected..
Any arrangement of the letters KJNN will yield a group with K and J.
Thus, to account for ALL OF THE WAYS that K and J can be selected, the result above must be multiplied by the number of ways to arrange the letters KJNN.
Number of ways to arrange 4 elements = 4!.
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 N's:
4!/2! = 12.
Multiplying the results above, we get:
P(K and J are selected) = 12 * 1/42 = 2/7.
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