Four fair dice D1-D2-D3 and D4- each having six faces 1-2-3-4-5-and 6 are rolled simultaneously- The probability nbsp-that D4 shows a no-appearing on one of D1-D2 and D3 is - Explain each step with a reason-

Total #160;no. #160;of #160;outcomes=64Favourable #160;cases #160;are: when #160;D4 #160;and #160;only #160;1 #160;of #160;D1D2D3 #160;s ...

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Standard VII

Mathematics

Outcome

Four fair dic...

Question

Four fair dice D1,D2,D3 and D4, each having six faces 1,2,3,4,5,and 6 are rolled simultaneously. The probability that D4 shows a no.appearing on one of D1,D2 and D3 is ? Explain each step with a reason.

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Solution

$Total no . of outcomes = 6^{4} Favourable cases are : - when D_{4} and only 1 of D_{1} D_{2} D_{3} shows same number or only 2 of D_{1} D_{2} D_{3} shows same number or all 3 of D_{1} D_{2} D_{3} shows same number . No . of favorable outcomes = {}^{6}C_{1} \times {}^{3}C_{1} \times 5 \times 5 + {}^{6}C_{1} \times {}^{3}C_{2} \times 5 + {}^{6}C_{1} \times {}^{3}C_{3} = {}^{6}C_{1} ({}^{3}C_{1} \times 5 \times 5 + {}^{3}C_{2} \times 5 + {}^{3}C_{3}) So, Required probability = \frac{No . of favorable outcomes}{Total no . of outcomes} = \frac{{}^{6}C_{1} ({}^{3}C_{1} \times 5 \times 5 + {}^{3}C_{2} \times 5 + {}^{3}C_{3})}{6^{4}} = \frac{6 \times (3 \times 25 + 3 \times 5 + 1)}{216 \times 6} = \frac{91}{216} ANS . . .$

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