Gaj81 wrote:5 boys and 4 girls are going for a movie. Andy and Sally are part of this group. For a restriction that no two boys or girls can sit beside each other, what is the probability of Andy sitting besides Sally?
Since no boy can sit next to another boy, the arrangement looks like this:
B-G-B-G-B-G-B-G-B
Case 1: Andy sits on EITHER END
P(Andy sits on either end) = 2/5. (Of the 5 boys' seats, 2 are on the end.)
P(Sally sits next to Andy) = 1/4. (Once Andy has been placed on either end, Sally must select the seat next to Andy, giving her only 1 good option from the 4 girls' seats.)
Since we want both events to happen, we MULTIPLY the fractions:
2/5 * 1/4 = 2/20 = 1/10.
Case 2: Andy sits in one of the MIDDLE SEATS
P(Andy sits in the middle) = 3/5. (Of the 5 boys' seats, 3 are in the middle.)
P(Sally sits next to Andy) = 2/4. (Once Andy has been placed, Sally must sit directly to the left or right of Andy, giving her 2 good options from the 4 girls' seats.)
Since we want both events to happen, we MULTIPLY the fractions:
3/5 * 2/4 = 6/20 = 3/10.
Since either Case 1 or Case 2 will yield a favorable outcome, we ADD the probabilities:
1/10 + 3/10 = 4/10 = [spoiler]2/5[/spoiler].
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