Question

A bag contains three white, two black and four red marbles. Four marbles are drawn at random
with replacement; calculate the probability that the sample contains just one white marble
given that it contains just one red marble.

Answer 1

Answer:

Hope this will help you

Step-by-step explanation:

You have two possible scenarios for the 2nd ball to be red

1st ball is red and 2nd ball is red

P= (2/9)*(1/8) = 2/72 = 1/36

1st ball is not red , 2nd ball is red

(7/9)*(2/8) = 14/72 = 7/36

so add this two probabilities

1/36 + 7/36 = 8/36 = 4/18 = 2/9

so 2/9 is the probability the 2nd ball is

Answer 2

Short Answer: 0.7

Long Answer:

Let A be the event that the first marble is white and B be the event that the second ball is black

We want the probability of A or B = A union B, ie P(A union B)

Let X’ denote the complement of X

So P(A union B) = 1 - P((A union B)’)

= 1 - P(A’ intersection B’)

Now A’ is the event that the first marble is NOT white, ie it is black

And similarly B’ is the event that the second marble is white

Thus A’ intersection B’ is the event that the first marble is black and second is white

So the number of ways this can happen is P(2, 1) x P(3, 1) (where P(n, r) is the number of ways of arranging r out of n objects)

= 2 x 3 = 6

But the total number of possibilities is P(5, 2) (as there are 5 balls in the beginning)

= 5 x 4 = 20

Thus P(A’ intersection B’) = 6/20 = 0.3

Hence, the required probability is 1 - 0.3 = 0.7