Bag I contains 3 white and 4 black balls, while Bag II contains 5 white
and 3 black balls. One ball is transferred at random from Bag I to Bag II
and then a ball is drawn at random from Bag II. The ball so drawn is
found to be white. Find the probability that the transferred ball is also
Answers
Answer:
sry I don't know bzhzhhzghs
bzjxjhdhjcbxgffagshjdjfkjcghs
xbjxjxjxn
jzjzhhzgxxvghhjBhzhhhsgggxgxgxg
kzjzhxhx
nxhhxhx
Answer:Let E
1
,E
2
,E
3
and A be events such that
E
1
= Both transferred ball from Bag I to bag II are red.
E
2
= Both transferred ball from Bag I to bag II are black.
E
3
= Out of two transferred ball one is red and other is black.
A = drawing a red ball from Bag II.
Here, P(
A
E
2
) is required.
Now, P(E
1
)=
7C
2
3C
2
=
2!1!
3!
×
7!
2!×5!
=
7
1
P(E
2
)=
7C
2
4C
2
=
2!2!
4!
×
7!
2!×5!
=
7
2
P(E
3
)=
7C
2
3C
1
×4C
1
=
7!
3!×4!
×
1!
2!×5!
=
7
4
P(
E
1
A
)=
11
6
,P(
E
2
A
)=
11
4
,P(
E
3
A
)=
11
5
Therefore, P(
A
E
2
)=
P(E
1
).(
E
1
A
)+P(E
2
).(
E
2
A
)+P(E
3
).(
E
3
A
)
P(E
2
).P(
E
2
A
)
aftersolving
=
77
6
+
77
8
+
77
20
77
8
=
77
8
×
34
77
=
17
4
Therefore, the probability that the transferred balls were both black =
17
4
Step-by-step explanation: