SOLUTION: We have 2 girls and 4 boys to be seated ina 6-seat row at the movie theater. What is the probability that the two people at each end of the row were both boys or both girls? Answer
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-> SOLUTION: We have 2 girls and 4 boys to be seated ina 6-seat row at the movie theater. What is the probability that the two people at each end of the row were both boys or both girls? Answer
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Question 391621: We have 2 girls and 4 boys to be seated ina 6-seat row at the movie theater. What is the probability that the two people at each end of the row were both boys or both girls? Answer should be in common fraction.
My approach to this problem:
Total possibilities of boys and girls is 4! * 2!
No of success outcomes: for Girls: 2! * 2!
for Boys : 4
p(both boys or both girls)= (4 + 2!*2!)/4! * 2!
Is my approach correct? Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! nPr=permutations of n things taken r at a time=n!/(n-r)!
.
a girl at either end:
2P2*4!=48 ways (only 2 possible arrangements for the girls and 4! for the boys.)
.
a boy at either end:
4P2 * 4!
=12*24
=288 ways
.
Probability=(48+288)/6!
=336/720
=7/15
.
Ed
