Question

2. On the Canadian Senate there are 105 members, 58 are Conservative and 47 are Liberal.

a) How many ways are there to select at least 2 Liberals for a sub-committee of 5 Senate members?

b) What is the probability of selecting 5 people for a sub-committee that contains 1 Liberals and 4 Conservatives?

c) What is the probability that no Liberals are selected in a sub-committee of 5 Senate members?

d) Suppose the 5 members in the sub-committee have rankings of President, Vice-President, Operations Officer, Press Secretary, and Treasurer. What is the probability of selecting all Liberals to these positions

Thanks! due tomorrow as well.

Answer #1

a) number of ways =P(at least 2 liberals)=N(2 liberal and 3 Conservative )+N(3 liberal and 2 Conservative )+N(4 liberal and 1 Conservative )+N(5 liberal and 0 Conservative )

=^{47}C_{2}*^{58}C_{3}
+^{47}C_{3}*^{58}C_{2}
+^{47}C_{4}*^{58}C_{1}
+^{47}C_{5}*^{58}C_{0}
=72037840

b) probability of selecting 5 people for a sub-committee that contains 1 Liberals and 4 Conservatives

=^{47}C_{1}*^{58}C_{4}
/^{105}C_{5} =47*424270/96560646 =0.206509

c)

probability that no Liberals are selected in a sub-committee of
5 Senate members=
^{47}C_{0}*^{58}C_{5}
/^{105}C_{5}

=1*4582116/96560646 =0.047453

d)

probability of selecting all Liberals to these positions
=^{47}C_{5} /^{105}C_{5}
=1533939/96560646 =0.015886

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