Question

Suppose we have four boxes A,B,C and D containing coloured marbles as given below :One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A? box B?, box C? Select the correct answer from above options

Answer 1

Probablity of selecting red ball from Box A,`P(R/A) = 1/10` Probablity of selecting red ball from Box B,`P(R/B) = 6/10 =3/5` Probablity of selecting red ball from Box C,`P(R/C) = 8/10 =4/5` Probablity of selecting red ball from Box D,`P(R/D) = 0/10 =0` Probablity of selected red ball is from Box A, `P(A/R)` can be given by, `P(A/R) = (P(R/A))/(P(R/A)+P(R/B)+P(R/C)+P(R/D))` `=(1/10)/(1/10+3/5+4/5) = 1/15` Probablity of selected red ball is from Box A, `P(A/R)` can be given by, `P(A/R) = (P(R/A))/(P(R/A)+P(R/B)+P(R/C)+P(R/D))` `=(1/10)/(1/10+3/5+4/5) = 1/15` Probablity of selected red ball is from Box B, `P(B/R)` can be given by, `P(B/R) = (P(R/B))/(P(R/A)+P(R/B)+P(R/C)+P(R/D))` `=(3/5)/(1/10+3/5+4/5) = 6/15=2/5` Probablity of selected red ball is from Box B, `P(B/R)` can be given by, `P(C/R) = (P(R/C))/(P(R/A)+P(R/B)+P(R/C)+P(R/D))` `=(4/5)/(1/10+3/5+4/5) = 8/15`