Jane is going into the next grade. Her friends, Sam and Lisa are attending the same school as her. She has 4 choices for a home room class. What are the chances at least one of her friends will end...

Jane is going into the next grade. Her friends, Sam and Lisa are attending the same school as her. She has 4 choices for a home room class. What are the chances at least one of her friends will end up in the same class as her?

Please help me figure out this problem as soon as possible. Thank you!

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

(1) Consider the probability that neither Sam nor Lisa have the same home room as Jane. The probability that Sam has a different home room is 3/4 as is the probability that Lisa has a different home room. Thus the probability that neither share a home room with Jane is (3/4)(3/4)=9/16.

The probability that at least one of Sam or Lisa sharing a home room with Jane is the complement of neither sharing a home room with her.

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Thus the probability is 1-9/16=7/16

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(2) An alternative approach:

The probability that Sam shares homerooms with Jane is 1/4.
The probability that Lisa shares homerooms with Jane is 1/4.
The probability that both share homerooms with Jane is (1/4)(1/4)=1/16

Thus the probability of at least one sharing her homeroom is 1/4+1/4-1/16=7/16. We subtract because the case that they are both in Jane's homeroom was counted twice.

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