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# Two dice are thrown. Find the probability that the number appeared have a sum 8 if it is known that second dice always exhibits 4.

SHAIK AASIF AHAMED
7 years ago
Hello student,
Second die always exhibits 4 so (1,4)(2,4)(3,4)(4,4)(5,4)(6,4)
as second die is always 4 and to sum 8 (4,4) is only possibility
P(B/A)=1/6
Thanks and Regards
Shaik Aasif
ANGEL MARY THOMAS
11 Points
3 years ago
P(S)=36
Event A :-Numbers appeared have a sum 8
{(2,6),(6,2),(3,5),(5,3),(4,4)}
Event B:-second die always exhibit 4
{(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)}
A INTERSECTION B={(4,4)}
P(B/A)= [P(B INTERSECTION A)]/P(B)
=[1/36 ]  / [6/36]
=1/6
Vikash
15 Points
2 years ago
By conditional probability:
We konw that P(A/B)={P(A intersection b)}/P(B).

Here P(a)=no of event in which sum appears 8
P(B)= second die exhibit 4

P(A)=(2,6)(3,5)(4,4)5,3)(6,2)

P(B)=(1,4)(2,4)(3,4)(4,4)(5,4)(6,4)

P(A intersection B)=(4,4)
•°• P(B/A)=(1/36)/(6/36)=1/6