NIMCET 2008 Question Paper with solutions

Question : 22 of 120

Marks: +1, -0

A bag contains 6 red and 4 green' balls. A fair dice is rolled and a number of balls equal to that appearing on the dice is chosen from the urn at random. The probability that all the balls selected are red is

${1}/{3}$
${3}/{10}$
${1}/{8}$
None of these

Validate

Solution:

Required probability

=P(1 R)+P(2 R)+P(3 R)+P(4 R)+P(5 R)+P(6 R)

= {1}/{6} × {6}/{10}+ {1}/{6} × {{ }^{6} C_{2}}/{{ }^{10} C_{2}}+ {1}/{6} × {{ }^{6} C_{3}}/{{ }^{10} C_{3}}+ {1}/{6} × {{ }^{6} C_{4}}/{{ }^{10} C_{4}} + {1}/{6} × {{ }^{6} C_{5}}/{{ }^{10} C_{5}}+ {1}/{6} × {{ }^{6} C_{6}}/{{ }^{10} C_{6}}

= {1}/{6}[ {3}/{5}+ {1}/{3}+ {1}/{6}+ {1}/{14}+ {1}/{42}+ {1}/{210}]

= {1}/{6} × {136}/{105}= {68}/{315}

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