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+2 votes
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One dice is thrown three times and the sum of the thrown numbers is 15.find the probability for which number 4 appears in first throw.
asked in Probability by Active (1.6k points)  
recategorized by | 171 views
1/18  ??

3 Answers

+4 votes

Total no of patterns where 15 as sum appears...

366,456,465,546,564,555,645,654,663,636

So probability = 2/10 = 1/5

answered by Veteran (22.8k points)  
edited by
I am getting same but ans is 1/18
Question already includes that sum is 15 ...

We need to find only how may cases it will give 15..among these 15 how many are favourable... That gives the probability..
@gabbar

n(s) = 6*6*6 =216

it is already given that 4 appears in 1st throw

so in 2nd and 3rd throw we can have (5,6) and (6,5) to make sum as 15

so n(E) ={(4,5,6),(4,6,5)}

p(E)=2/216

      =1/105............ryt??
Here Sample Space is 10 (sum of the thrown number is 15).
+1 vote
its 2/10 ie 1/5
answered by (427 points)  
0 votes
ans = P( number in 1st dice throw is 4 & sum of numbers in 2nd and 3rd dice throw is 11 ) / P(number in 1st dice throw is 4)

= (1/6 * 2/36 ) / ( 1/6 )

=1/18
answered by Veteran (15.1k points)  


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