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What is the probability distribution for X?

A box has five tickets numbered 1, 2, 3, 4, and 5. Two are to be randomly selected without replacement. Let X be the number of occurrences of either 4 or 5. For example, if 2 and 5 are selected, then X=1. If both 4 and 5 are selected, then X=2.

A box has five tickets numbered 1, 2, 3, 4, and 5. Two are to be randomly selected without replacement. Let X be the number of occurrences of either 4 or 5. For example, if 2 and 5 are selected, then X=1. If both 4 and 5 are selected, then X=2.

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Mar 20, 2018

Answer:

#P(x) = {(1/10, x = 2),(6/10,x=1),(3/10,x=0):}#

Explanation:

We are choosing 2 tickets from 5 tickets, therefore, the number of combinations are:

#( (n),(k) ) = (n!)/(k!(n-k!)#

where #n = 5# and #k = 2#

#( (5),(2) ) = (5!)/(2!(3!)) = 10#

Counting all of the possible outcomes:

#P(2) = (4,5) = 1/10#
#P(1)= (1,4), (2,4),(3,4),(1,5), (2,5),(3,5) = 6/10#
#P(0) = (1,2),(1,3),(2,3)= 3/10#

Please observe that #P(2)+P(1) + P(0) = 1#; this is a good indicator that the distribution is valid:

#P(x) = {(1/10, x = 2),(6/10,x=1),(3/10,x=0):}#

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