# Three dices D1,D2,D3 are thrown. Find the probability that two numbers of two dices matches only.

We have to find the probability that two numbers of two dice match :

1)$Thenumberofcasesinwhichallthethreedicegivethesamenumberare:\phantom{\rule{0ex}{0ex}}(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)=6cases$

2)$Thenumberofcasesinwhichalldicegivedifferentnumbers=6\times 5\times 4=120\phantom{\rule{0ex}{0ex}}$

(This is because when the first die is thrown,any 6 numbers are but when the second die is thrown 5 possible number are there because in this case the second die cant show the same number as the first die and similarly fir the third die 4 numbers are possible)

3)$Sothecasesinwhichtwodiematch=Totalcases-casesinwhichalldieshowsamenumber-casesinwhichalldieshowdifferentnumbers\phantom{\rule{0ex}{0ex}}=216-120-6=90$

SO the required probability=$\frac{90}{216}=\frac{5}{12}$

Regards

(This is because when the first die is thrown,any 6 numbers are but when the second die is thrown 5 possible number are there because in this case the second die cant show the same number as the first die and similarly fir the third die 4 numbers are possible)

3)$Sothecasesinwhichtwodiematch=Totalcases-casesinwhichalldieshowsamenumber-casesinwhichalldieshowdifferentnumbers\phantom{\rule{0ex}{0ex}}=216-120-6=90$

SO the required probability=$\frac{90}{216}=\frac{5}{12}$

Regards

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