A box has 7 white and 3 red balls, while
another box has 4 white and 3 red balls. One
ball is randomly selected from the first box
and is transferred to the second box. After
that, one ball is randomly drawn from the
second box. Find the probability that it is red.
a) 0.41 b) 0.46 c) 0.49 d) 0.48
Answers
solve this question, We have to use Bayes Theorem to get reverse probability.
Let first ball drawn is red and second ball drawn is white.
Probability of choosing a box at random =
3
1
.
probabilty of drawing a red ball from box−1=p(R∣B
1
)=
3
1
probabilty of drawing a red ball from box−2=p(R∣B
2
)=
2
1
probabilty of drawing a red ball from box−3=(R∣B
3
)=
6
1
[12/2, 4:14 PM] Mera Baccha: probability of getting the ball from box−2 if red ball is selected is given by
P(B
2
∣R)=
P(B
1
)×p(R∣B
1
)+P(B
2
)×p(R∣B
2
)+P(B
3
)×p(R∣B
3
)
P(B
2
)×p(R∣B
2
)
=
3
1
×
3
1
+
3
1
×
2
1
+
3
1
×
6
1
3
1
×
2
1
=
9
1
+
6
1
+
18
1
6
1
=
18
2+3+1
6
1
[12/2, 4:16 PM] Mera Baccha: 1
=
6
1
×
6
18
=
2
1
......(1)
Now let us consider white ball is drawn second time
probabilty of drawing a white ball from box−1=p(W∣B
1
)=
6
1
probabilty of drawing a white ball from box−2=p(W∣B
2
)=
5
2
( note:- one ball is taken out already from box−2)
probabilty of drawing a white ball from box−3=p(W∣B
3
)=
2
1
probability of getting the ball from box−2 if white ball is selected is given by
P(B
2
∣W)=
P(B
1
)×p(W∣B
1
)+P(B
2
)×p(W∣B
2
)+P(B
3
)×p(W∣B
3
)
P(B
2
)×p(W∣B
2
)
=
3
1
×
6
1
+
3
1
×
5
2
+
3
1
×
2
1
3
1
×
5
2
=
18
1
+
15
2
+
6
1
15
2
Hence probability of getting the balls from box−2 if red ball selected first time and white ball second time =
2
1
×
8
3
=
16
3
...................(3)$$
Now we consider the event, first selected ball is white and the second one red
probabilty of drawing a white ball from box−1=p(W∣B
1
)=
6
1
probabilty of drawing a white ball from box−2=p(W∣B
2
)=
3
1
probabilty of drawing a white ball from box−3=p(W∣B
3
)=
2
1
probability of getting the ball from box−2 if white ball is selected is given by
P(B
2
∣W)=
P(B
1
)×p(W∣B
1
)+P(B
2
)×p(W∣B
2
)+P(B
3
)×p(W∣B
3
)
P(B
2
)×p(W∣B
2
)
=
3
1
×
6
1
+
3
1
×
3
1
+
3
1
×
2
1
3
1
×
3
1
=
18
1
+
9
1
+
6
1
9
1
[12/2, 4:17 PM] Mera Baccha: P(B
1
)×p(W∣B
1
)+P(B
2
)×p(W∣B
2
)+P(B
3
)×p(W∣B
3
)
P(B
2
)×p(W∣B
2
)
=
3
1
×
6
1
+
3
1
×
3
1
+
3
1
×
2
1
3
1
×
3
1
=
18
1
+
9
1
+
6
1
9
1
=
3
1
......(4)
Now let us consider red ball is drawn second time
probabilty of drawing a red ball from box−1=p(R∣B
1
)=
3
1
probabilty of drawing a red ball from box−2=p(R∣B
2
)=
5
3
probabilty of drawing a red ball from box−3=p(R∣B
3
)=
6
1
probability of getting the ball from box−2 if red ball is selected second time is given by
P(B
2
∣R)=
P(B
1
)×p(R∣B
1
)+P(B
2
)×p(R∣B
2
)+P(B
3
)×p(R∣B
3
)
P(B
2
)×p(R∣B
2
)
=
3
1
×
3
1
+
3
1
×
5
1
+
3
1
×
6
1
3
1
×
5
3
=
9
1
+
5
1
+
18
1
15
3
=
11
6
......(5)
Hence probability of getting the balls from box−2 if white ball is selected first time and red ball is selected second time =
3
1
×
11
6
=
11
2
...................(6)
Using (3) and (6), probability of getting two balls from box−2 if one of then red and other one is white =
16
3
+
11
2
=
176
65
=0.37