PROBALITY-3

'manpreet singh wrote:Q.1) In a bag of marbles, there are 3 red, 2 white, and 5 blue. If Bob takes 2 marbles out of
the bag, what is the probability that he will have one white and one blue marble?

P(1st marble is white) = 2/10. (Of the 10 marbles, 2 are white.)
P(2nd marble is blue) = 5/9. (Of the 9 remaining marbles, 5 are blue.)
Since we want both events to happen, we multiply the probabilities:
2/10 * 5/9 = 1/9.
Since a good outcome will be achieved if the order of the colors is reversed, we multiply by 2:
2 * 1/9 = 2/9.

I dont understand why order is considered while taking out the probability.

ritind wrote:I dont understand why order is considered while taking out the probability.

In the problem above, there is a SINGLE EVENT: the selecting of two marbles.
We need to account for ALL OF THE WAYS that this single event could yield a white marble and a blue marble.
There are TWO WAYS that this single event could yield a favorable outcome:
Case 1: If the first marble selected is white, and the second marble selected is blue.
Case 2: If the first marble selected is blue, and the second marble selected is white.
The probability of a favorable outcome is THE SAME in each case: 1/9.
Thus, the probability that this SINGLE EVENT OF SELECTING TWO MARBLES will yield a white marble and a blue marble = 2 * 1/9 = 2/9.

One of the ways to solve this problem will be to calculate total number of combinations of having 2 marbles out of 10.
10!/2!(10-2)!=45 so there are 45 possible combinations

Now we need to find number of combinations when we have one white and one blue marble.
one white marble - 2!/1!(2-1)!=2
one blue marble - 5!/1!(5-1)!=5
So the total combinations of having 1 white and 1 blue combinations = 2*5=10

Probability of having 1 white and 1 blue = 10/45=2/9

Hope that helps!