IIT JEE , AIEEE Forum - Mathematics , Algebra - probability-tricky ques

warrior

there are 2 unbiased dies such that probability of getting even no. is twice the probability of getting odd no. Two such dies are thrown together. probability that product of numbers on them is 4=

ans: 8/(8!)

warrior

need explanation

prateek2292

hey
the probability of getting even no. & odd no. as mentioned by u, in that case even & odd no.'s are on both the dies or on any.

PLEASE LET ME KNOW.........

warrior

the given condition applies to both dies
product=4 implies
2 , 2
1 ,4
4 , 1 i am not getting that ans

akhil_o

let prob for 1 even no. to occur=e
and 1 odd=o
3o+3e=1
and
e=2o
so
9o=1,
o=1/9
and e=2/9

probabilities for 4:
1,4 4,1 and 2,2 gives 4

so
(1/9)(2/9)2 + (2/9)(2/9) gives required prob
=4/81+4/81
=8/81

PS note its 8/81 not 8/8!

sinjan.j

wait.. i'll answer

sinjan.j

check if this answers your question or not

first of all if probability of even number coming is twice that of odd
the cunning dies is no way unbiased .
'unbiased; is not an honorary title for dies.
say prob of getting (2,4,6)=p1
while that of (1,3,5)=p2
the sum(pi)=1=> p2=1/9
4 could occur as (2,2)U (4,1) U(1,4)=p1 p1+2p1p2=8/9^2 prima facie

see if this answers.. if not then ask the doubt and nudge me..!!!

magiclko

akhil and sinjan , both are correct!!!

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