3 couples and 2 single individuals have been invited to aninvestment seminar and

Question

3 couples and 2 single individuals have been invited to aninvestment seminar and have agreed to attend. Suppose theprobability that any particular couple or individual arrives lateis .4 (A couple will travel together, so either both people arelate or else both are). Assume different individuals or couples arelate independently of each other. Let X = the number of people whoarrive late.

a) Determine the probability mass function of X.
b) Obtain the cumulative distribution function of X and use it tocalculate P(2 <= x <= 6)

I think part a) is pretty straight forward, since the probabilityof two independent events occurring is just p(event one) * p(eventtwo)

I'm struggling with part b) however. We didn't really go overcumulative distribution functions in class. I'm also wondering howto solve for the probability at most 6 people can be late when only5 groups are attending...

Thanks for any help!

Explanation / Answer

The definition of X is : "Let X = the number of peoplewho arrive late."    Here there are in total 8 persons, not 5.    a) So, P( X = 0 ) = P( nobody is late ) = ( 1 - 0.4 )5 .                P( X = 1 ) = P ( only one of the two individuals islate ) = 2 (0.4) ( 1 - 0.4 )4                P( X = 2 ) = P ( both the individuals are late orexactly one couple is late )       = (0.4)2 ( 1 - 0.4 )3    + 3 (0.4) (1 - 0.4)4                P( X = 3 ) = P ( exactly one of the three couples andone of the two individuals are late )       = (2x 3 ) (0.4)2 ( 1 - 0.4 )3 = 6(0.4)2 ( 1 - 0.4 )3                P( X = 4 ) = P ( exactly two of the three couples arelate or both individuals and a couple are late )       = 3(0.4)2 ( 1 - 0.4 )3 +3(0.4)3 ( 1 - 0.4 )2                P( X = 5 ) = P ( exactly two of the three couples andone of the two individuals are late )       = (3 x 2 )(0.4)3 ( 1 - 0.4 )2 = 6(0.4)3 ( 1 - 0.4 )2                P( X = 6 ) = P ( all the three couples are late or oneof the couple and both the individuals are late )       =(0.4)3 ( 1 - 0.4 )2 + 3(0.4)3 ( 1 - 0.4 )2                P( X = 7 ) = P ( all the three couples are late and oneof the individuals are late )       = 2(0.4)4 ( 1 - 0.4 )1                P( X = 8 ) = P ( all are late )       = (0.4)5

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