Probability Problems Based on Combinations

Question 1. An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that
(i) both the balls are red
(ii) one ball is red, the other is black
(iii) one ball is white

Solution:

Number of white balls = 7
Number of black balls = 5
Number of red balls = 3
Total number of balls = 7 + 5 + 3 = 15
Total number of ways of drawing 2 balls out of 15 balls = $^{15}C_{2}$
(i) Number of favourable cases for both the balls to be red = $^{3}C_{2}$
Therefore, the required probability = $\frac{^{3}C_{2}}{^{15}C_{2}} = \frac{3 \times 2}{15 \times 14} = \frac{1}{35}$
(ii) Number of favourable ways for one red ball and other black ball = $^{3}C_{1} \times ^{5}C_{1} = 3 \times 5 = 15$
Therefore, $P(\text{Drawing one red ball and the other black ball})$
= $\frac{^{3}C_{1} \times ^{5}C_{1}}{^{15}C_{2}} = \frac{15}{^{15}C_{2}} = \frac{15}{\frac{15 \times 14}{2 \times 1}} = \frac{1}{7}$
(iii) One ball is white.
Number of such favourable ways = $^{7}C_{1} \times ^{8}C_{1} = 7 \times 8 = 56$
Therefore, $P(\text{Drawing one white ball})$
= $\frac{56}{^{15}C_{2}} = \frac{56}{\frac{15 \times 14}{2 \times 1}} = \frac{8}{15}$

Question 2. Three cards are drawn at random from a pack of well shuffled deck of 52 cards. Find the probability that
(i) all the three cards are of the same suit.
(ii) one is king the other is quenn and third is a jack.

Solution:

Number of ways of drawing three cards from a well shuffled pack of 52 cards.
$n(S) = {{}^{52}C_3} = \frac{52!}{3! \cdot 49!} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100$

$\text{(i) Let } E \text{ = Event of getting all the three cards of the same suit.}$
$\text{There are four suits; namely, clubs, spades, hearts, and diamonds, each having 13 cards.}$
$n(E) = \text{number of ways of getting all the three cards of the same suit}$
$= \text{number of ways of getting three clubs or three spades or three hearts or three diamonds}$
$= {{}^{13}C_3} + {{}^{13}C_3} + {{}^{13}C_3} + {{}^{13}C_3} = 4 \times {{}^{13}C_3}$
$= 4 \times \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 1144$
$P(\text{getting three cards of same suit}) = P(E) = \frac{n(E)}{n(S)} = \frac{1144}{22100} = \frac{22}{425}$

$\text{(ii) Let } F = \text{Event of getting 1 king, 1 queen, and 1 jack.}$
$\text{We know that there are four kings, four queens, and four jacks.}$
$n(F) = \text{number of ways of selecting one king, one queen, and one jack}$
$= {{}^4C_1} \times {{}^4C_1} \times {{}^4C_1} = 4 \times 4 \times 4 = 64$
$P(\text{getting one king, one queen, and one jack}) = P(F) = \frac{n(F)}{n(S)} = \frac{64}{22100} = \frac{16}{5525}$

Question 3. The bag X contains 5 red and 3 green balls and bag Y contains 3 red and 5 green balls. One ball is drawn from bag X and two from bag Y. Find the probability that out of the three ball drawn two are red and one is green.

Solution:

Question 4. A bag continas 30 tickets numbered from 1 to 30. Five tickets are drawn at random and arranged in ascending order. Find the probability that third number is 20.

Solution:

A bag contains 10 red, 3 green balls, while another bag contains 3 red and 5 green balls. Two balls are drawn from first bag and put into the second bag and then a ball is drawn from the later. Find the probability that it is a red ball.

Solution: