A box contains 10 balls of which 6 are red (R) and 4 are white (W) . Three balls are taken from the box at random, one after the other. The probability that the 3 balls drawn are in the order RRW is equal to,
Pr=6/10*5/9*4/8= ?
To find the probability that the three balls drawn are in the order RRW, we need to consider the total number of possible outcomes and the number of favorable outcomes.
Total number of possible outcomes:
When three balls are drawn from a box of 10 balls without replacement, the number of possible outcomes can be calculated using combinations. This is denoted as "nCr" and can be calculated as:
nCr = n! / (r!*(n-r)!)
In this case, we want to find the number of ways we can choose 3 balls out of 10 without replacement, so n = 10 and r = 3.
Therefore, the total number of possible outcomes is:
10C3 = 10! / (3!*(10-3)!) = 10! / (3! * 7!) = (10*9*8) / (3*2*1) = 120
Number of favorable outcomes:
In order for the three balls to be in the order RRW, the first two balls must be red (R) and the third ball must be white (W). We have 6 red balls and 4 white balls in the box.
The probability of the first ball being red is 6/10, as there are 6 red balls out of 10 in total.
Once the first ball is red, there are only 5 red balls left out of 9 remaining balls, so the probability of the second ball being red is 5/9.
Finally, there are 4 white balls remaining out of 8 remaining balls, so the probability of the third ball being white is 4/8.
Therefore, the number of favorable outcomes is:
(6/10) * (5/9) * (4/8) = 120/720 = 1/12
Probability that the three balls drawn are in the order RRW:
The probability is given by the number of favorable outcomes divided by the number of total possible outcomes:
P(RRW) = favorable outcomes / total outcomes = (1/12) / (120/720) = 6/120 = 1/20
So, the probability that the 3 balls drawn are in the order RRW is 1/20.