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- A bag contains 5 red and 3 green balls.
Question 1:
A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag.Find the probability that one ball is red and one is green.
A. 19/20
B. 17/20
C. 8/10
D. 21/40
Feedback Let A be the event that ball selected from the first bag is red and ball selected from the second bag is green.
Let B be the event that ball selected from the first bag is green and ball selected from the second bag is red.
P(A)=(5/8)×(6/10)=3/8 and P(B)=(3/8)×(4/10) =3/20
Hence, required probability,
=P(A)+P(B)
=3/8+3/20
=21/40
Let B be the event that ball selected from the first bag is green and ball selected from the second bag is red.
P(A)=(5/8)×(6/10)=3/8 and P(B)=(3/8)×(4/10) =3/20
Hence, required probability,
=P(A)+P(B)
=3/8+3/20
=21/40
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