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?
OA-2/5
Probability
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Since no boy can sit next to another boy, the arrangement looks like this: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?
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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Many probability questions can be solved intuitively rather than with long calculations. Let's try to do so with this problem!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?
OA-2/5
We can get off to a great start by flipping around the question: what's the probability that Sally sits beside Andy?
We know that it has to go BGBGBGBGB. We can see that each girl sits beside 2 of the boys. Since there are 5 boys, there's therefore a 2/5 chance that Sally sits beside Andy - done!
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Since the 5 boys must be separated, the arrangement looks like this:Milovan wrote:Just from curiosity, how would you solve this one using permutations calculation?
B-G-B-G-B-G-B-G-B.
Total possible arrangements:
Number of ways to arrange the 5 boys = 5!.
Number of ways to arrange the 4 girls = 4!.
To combine these options, we multiply:
5!*4!.
Good arrangements:
In a good arrangement, Andy is next to Sally.
Number of options for Sally = 4. (Any of the 4 even-numbered seats.)
Number of options for Andy = 2. (To the left or right of Sally.)
Number of ways to arrange the remaining 3 girls = 3!.
Number of ways to arrange the remaining 4 boys = 4!.
To combine these options, we multiply:
4*2*3!*4!
P(Andy is next to Sally) = (good arrangements)/(total possible arrangements) = (4*2*3!*4!)/(5!*4!) = 2/5.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
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