Two baised dice are thrown together. For the first die `P(6)=1/2` , other scores being equally likely while for the second die, `P(1)=2/5` and other scores are equally likely. Find the probability distribution of the number of ones seen.

Explanation

For first die, `P(6)=1/2andP(6')=1/2`
`rArrP(1)+P(2)+P(3)+P(4)+P(5)=1/2`
`rArr P(1)=1/10andP(1')=9/10` `[thereforeP(1)=P(2)=P(3)=P(4)=P(5)]`
For second die, `P(1)=2/5and P(1')=1-2/5=3/5`
Let X= Number of one's seen
For X = 0, `P(X=0)=P(1')cdotP(1')+P(1')cdotP(1')=9/10cdot2/5+1/10cdot3/5`
`=18/50+3/50=21/50=0.42`
`P(X=2)=P(1)cdotP(1)=1/10cdot2/5 =2/50=0.04`
Hence, the required probablility distribution is as below