harshjoshi91 wrote:
A row of seats in a movie hall contains 10 seats. 3 Girls and 7 boys need to occupy those seats. What is the probability that no two girls will sit together?
ANSWER:
Here is a solution based on direct calculations of probabilities:
The probability of two girls sitting together is \(9\cdot\frac{3}{10}\cdot\frac{2}{9}=\frac{3}{5}.\)
When sitting 2 girls together - GG - 3 out of 10 choices for the first girl, then 2 out of 9 for the second girl.
The factor 9 accounts for the 9 possibilities where to place the two girls in the row of 10 seats:
GG********
*GG*******
...
********GG
The probability of all three girls sitting together is \(8\cdot\frac{3}{10}\cdot\frac{2}{9}\cdot\frac{1}{8}=\frac{1}{15}.\)
When sitting all 3 girls together - GGG - 3 out of 10 choices for the first girl, then 2 out of 9 for the second girl, and 1 out of 8 for the third girl.
The factor 8 accounts for the 8 possibilities where to place the three girls in the row of 10 seats:
GGG*******
*GGG******
...
*******GGG
Now we should be careful: when computing the probability for 2 girls sitting together, we didn't say anything about the third girl.
Those sitting arrangements include also cases of all three girls sitting together, and maybe more than once.
In fact, we can determine that we included all the cases when 3 girls sit together twice.
For any arrangement of two girls together, say GG, and the third girl g sitting with them, we also considered the arrangements gGG and GGg when placed on the same three seats, like
GGg*******
gGG*******
...
***GGg****
***gGG****
...
Therefore, for the final probability we get \(1-\frac{3}{5}+\frac{1}{15}=\frac{7}{15}.\)
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