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A family of 6 boys and 3 girls. Mother wanted to choose 4 of them to prepare dinner:
What is the probability of choosing a boy to prepare tea, a boy to cook, and two girls
to process Table?
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The total number of ways to assign jobs is
= 9*8*21 = 1512.
(9 ways to fill the first position; 8 ways to fill the second position;
and ways to fill the 3-rd and 4-th positions with possible transposition there).
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| Now let's calculate the number of ways |
| to assign jobs as described. |
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The number of ways to select one boy from 6 boys to prepare tea is 6.
The number of ways to select 1 boy from remaining 5 boys to cook is 5.
The number of ways to select 2 girls from 3 girls is 3.
So, the number of ways to form a proper team as described is 6*5*3 = 90.
The probability under the problem's question is
number of proper ways 90 5
P = ------------------------- = ----- = ---- = 0.05952 (rounded). ANSWER
total number of ways 1512 84
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| Another way to solve is to write |
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P = = = (the same answer).
Here is the probability to select one of 6 boys from 9 kids to prepare tea.
is the probability to select one of remaining 5 boys from remaining 8 kids to cook.
is the probability to select two of 3 girls from remaining 7 kids to process table.
Solved in two ways, for your better understanding.