In an AP first term is 2 and the sum of the first five terms is

Q) A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that:
(i) One is red and two are white
(ii) Two are blue and one is red
(iii) One is red

[Practice Paper 1, 2023-24, Dir of Edu, GNCT of Delhi]

Ans:

Total number of balls = 6 + 4 + 8 = 18 balls
Since 3 balls are being pulled out at random,
Hence, total possible number of ways of pulling 3 balls out of 18 balls
= $^{18}C_3$
= $\frac{18!}{3! 15!}$
= $\frac{(18) (17) (16) 15!}{(6) 15!}$
= $\frac{(18) (17) (16) \cancel{15!}}{(6) \cancel{15!}}$
= $\frac{(18) (17) (16)}{6}$
= 3 x 17 x 16
= 816

Hence, total possible number of outcomes = 816

(i) probability that One is red and two are white:
Let’s first calculate the favorable number of chances of getting the 1 red and 2 white:
There are 6 red balls and 1 has to be pulled out. Since each of the 6 balls have equal chances of being pulled out, hence there are $^6C_1$ chances i.e. $\frac{6!}{1!5!}$ = 6
Next, 2 white balls are pulled from 4 white balls and each ball has equal chance of being pulled out, hence there are $^4C_2$ chances i.e $\frac{4!}{2! 2!}$ = 6

Since both events are mutually exclusive (i.e. have no dependency on each other) Hence, total number of favorable chances = 6 x 6 = 36

Hence probability = $\frac{favorable~outcomes}{total~outcomes}$ = $\frac{36}{816}$
= $\frac{3}{68}$

(ii) Two are blue and one is red
Let’s first calculate the favorable number of chances of getting the 2 blue and 1 red:
First, 2 blue balls are pulled from 8 blue balls and each ball has equal chance of being pulled out, hence there are $^8C_2$ chances i.e $\frac{8!}{2! 6!}$ = 28
Next, there are 6 red balls and 1 has to be pulled out. Since each of the 6 balls have equal chances of being pulled out, hence there are $^6C_1$ chances i.e. $\frac{6!}{1!5!}$ = 6

Hence, total number of favorable chances = 28 x 6 = 168

Since probability = $\frac{favorable~outcomes}{total~outcomes}$ = $\frac{168}{816}$
= $\frac{7}{34}$

(iii) One is red
If one ball is red, then there is possibility that among other two balls, both are white or both are blue or one is white and one is blue
Hence combinations will be: 1R2W or 1R2B or 1R1W1B

First for 1R2W:
There are 6 red balls and 1 has to be pulled out. Since each of the 6 balls have equal chances of being pulled out, hence there are $^6C_1$ chances i.e. $\frac{6!}{1!5!}$ = 6

Next, 2 white balls are pulled from 4 white balls and each ball has equal chance of being pulled out, hence there are $^4C_2$ chances i.e $\frac{4!}{2! 2!}$ = 6

Hence chances for 1R2W combination = 6 x 6 = 36

Next for 1R2B:
There are 6 red balls and 1 has to be pulled out. Since each of the 6 balls have equal chances of being pulled out, hence there are $^6C_1$ chances i.e. $\frac{6!}{1!5!}$ = 6

Next, 2 blue balls are pulled from 8 blue balls and each ball has equal chance of being pulled out, hence there are $^8C_2$ chances i.e $\frac{8!}{2! 6!}$ = 28

Hence chances for 1R2B combination = 6 x 28 = 168

Next for 1R1W1B:
There are 6 red balls and 1 has to be pulled out. Since each of the 6 balls have equal chances of being pulled out, hence there are $^6C_1$ chances i.e. $\frac{6!}{1!5!}$ = 6

Next, 1 white ball is pulled from 4 white balls and each ball has equal chance of being pulled out, hence there are $^4C_1$ chances i.e $\frac{4!}{1! 3!}$ = 4

Next, 1 blue ball is pulled from 8 blue balls and each ball has equal chance of being pulled out, hence there are $^8C_1$ chances i.e $\frac{8!}{1! 7!}$ = 8