Sam and jessica are invited to a dance .if there are 7 women and 7 womenin total at the dance, and one woman and one man are chosen to lead the dance , what is the probability that sam and jessica will not be the pair chosen to lead the dance?
1/49 1/7 6/7 47/49 48/49
I have solved this Q ,so obviously i am not looking for answer but my doubt is why can't we use the following approach....
Since we dont want jessica and sam to lead the pair ,so we can choose the pair in 6/7 * 6/7 ways
which is 36 /49 but this is wrong .where i am wrong....
probability doubt
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Hi,if there are 7 women and 7 womenin total at the dance,
I guess it is 7 men and 7 women.
Coming to the problem,
You calculation contains all pairs in which neither Sam nor Jessica is chosen. But, the question asks all pairs except the pair of Sam and Jessica. There is a difference, right?
For instance, you are leaving the cases in which Sam is chosen and any of the other 6 girls except Jessica is chosen and the cases in which Jessica is chosen and any of the other 6 boys except Sam is chosen.
This adds (1/7)*(6/7) + (6/7)*(1/7) to the already calculated 36/49.
So, answer is 36/49+6/49+6/49 = 48/49.
A simpler way would be removing the case of choosing Sam and Jessica.
Sam and Jessica can be chosen as a pair with probability (1/7)*(1/7) = 1/49.
So, probability that sam and jessica will not be the pair chosen = 11/49 = 48/49.
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise