Let #R_1# be the Event that a Red Marble is obtained in the #1^(st)#
attempt, and, #R_2, the similar event.
#"Then, Reqd. Prob.="P(R_1nnR_2)...................(ast)#
For #P(R_1),# there are #6 (R)+5 (G)+4 (Y)=15# marbles in the bag, out of which, #1# can be picked in #15# ways.
There are #6# Red marbles in the bag, out which #1# can be picked in #6# ways.
Therefore, #P(R_1)=6/15.................................(1)#
For #P(R_2)#, let us remember that, while selecting the #2^(nd)# marble from the bag, the #1^(st)# selected marble is not to be replaced back in the bag. As a result of this, now, in the bag, there are #(6-1=5) (R)+5 (G)+4 (Y)=14# marbles in the bag.
Thus, the Probability of the Event #R_2# (i.e., of picking #2^(nd)# Red marble), knowing that the Event #R_1# has already occurred (i.e., a Red marble has been picked in the #1^(st)# trial) is,
#P(R_2/R_1)=5/14........................................(2)#
#"Therefore, the Reqd. Prob.="P(R_1nnR_2)#
#=P(R_1)*P(R_2/R_1)=6/15*5/14#
#rArr "the Reqd. Prob.="1/7#.
Enjoy Maths.!
