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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.
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1/18  ??

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 (23.2k 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)
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 (15.4k points)
Why you have divided by P(no. in 1st dice throw is 4).

As condition is given as sum is 15, and we divide with the probability of given condition in conditional probability, right?
I am sorry. It should be

Ans= p (4 on first dice and sum of other 2 dice is 11)

/{ p (4 on first dice and sum of other 2 dice is 11)+

p (First dice has other than 4 and sum of all 3 is 15)}

=2/10

=1/5