CAT 2025 Lesson : Probability - Dependent Events

A bag contains $7$ red balls and $5$ black balls. What is the probability that the two balls drawn at random are

(I) both black when the balls are drawn with replacement.
(II) both black when the balls are drawn without replacement.
(III) one red and one black when the balls are drawn with replacement.
(IV) one red and one black when the balls are drawn without replacement.

Solution

When a ball is drawn and then placed back in the bag before the next ball is drawn, then this is a case where the balls are being drawn with replacement. The two balls being such drawn are two independent events as the first drawn ball does not impact the probability of the second ball. (Cases I and III)

When the balls are drawn without replacement, the drawn balls are not replaced in the bag. Therefore, probability for the second ball will be impacted by the colour of the first drawn ball. (Cases II and IV)

Case I: Probability of $2$ black balls drawn with replacement $= \dfrac{5}{12} \times \dfrac{5}{12} = \dfrac{\bm{25}}{\bm{144}}$

Case II: Probability of first ball being black $= \dfrac{5}{12}$

Now, the first ball, coloured black, has been drawn. So, we are left with 4 black ball in a total of $11$ balls.

Probability of second ball being black $= \dfrac{4}{11}$

Probability of $2$ black balls drawn without replacement $= \dfrac{5}{12} \times \dfrac{4}{11} = \dfrac{\bm{5}}{\bm{33}}$

Case III: Probability of red and black balls drawn in that order with replacement $= \dfrac{5}{12} \times \dfrac{7}{12} = \dfrac{\bm{35}}{\bm{144}}$

Probability of black and red balls drawn in that order with replacement$= \dfrac{7}{12} \times \dfrac{5}{12} = \dfrac{\bm{35}}{\bm{144}}$

Probability of 1 red and 1 black balls drawn with replacement $= \dfrac{35}{144} + \dfrac{35}{144} = \dfrac{\bm{35}}{\bm{72}}$

Case IV: There are again 2 possibilities here – the first and second balls are red and black or black and red respectively.

Probability of 1 red and 1 black balls drawn without replacement $= \dfrac{5}{12} \times \dfrac{7}{11} + \dfrac{7}{11} \times \dfrac{5}{11} = \dfrac{\bm{35}}{\bm{66}}$

Alternative Methods for Cases II and IV

Case II: We can use the combinations or selections formula.

Probability of 2 black balls drawn without replacement $= \dfrac{ \mathrm{Ways to select 2 black balls}}{\mathrm{Ways to select 2 balls}} = \dfrac{^{5}C_2}{^{12}C_2}= \dfrac{10}{66}= \dfrac{\bm{5}}{\bm{33}}$

Case IV: Probability of drawing 1 black and 1 red ball without replacement $= \dfrac{^{5}C_1 \times ^{7}C_1} {^{12}C_2}= \dfrac{\bm{35}}{\bm{66}}$

Answer: (I) $\dfrac{25}{144} ;$ (II) $\dfrac{5}{33} ;$    (III)$\dfrac{35}{72} ;$    (IV) $\dfrac{35}{66}$

From a standard deck of 52 cards, 2 cards are drawn at random one after the other without replacement. What is the probability that
(I) Both the cards are red?
(II) one of the cards is spades while the other is clubs?
(III) the first card is Jack and the second card is King?
(IV) the two cards are Jack and King?
(V) the cards are a queen and a red-coloured King?

Solution

As shown in the previous example for Cases II and IV, we shall use the normal dependent probability method and selections method to solve all the cases.

Cases	Normal Method	Selections Method
Case I: There are 26 red cards	$\dfrac{26}{52} \times \dfrac{25}{51} =\dfrac{25}{102}$	$\dfrac{^{26}C_2}{^{52}C_2} = \dfrac{25}{102}$
Case II: There are 13 spades and 13 clubs	Note: We add the probabilities of 1st spade & 2nd clubs and that of 1st clubs & 2nd spade	$\dfrac{^{13}C_1 \times ^{13}C_1}{^{52}C_2} = \dfrac{13}{102}$
Case III: There are 4 Jacks and 4 Kings (Order matters)	$\dfrac{4}{52} \times \dfrac{4}{51} =\dfrac{4}{663}$	$\dfrac{1}{2} \times \dfrac{^{4}C_1 \times ^{4}C_1}{^{52}C_2} = \dfrac{4}{663}$ Note: The selection probability is for 2 cards being Jack and King in any order. However, as the 1st and 2nd have to be Jack and King respectively, we multiply by
Case IV: There are 4 Jacks and 4 Kings (Order does not matter)	$\dfrac{4}{52} \times \dfrac{4}{51} + \dfrac{4}{52} \times \dfrac{4}{51} =\dfrac{8}{663}$	$\dfrac{^{4}C_1 \times ^{4}C_1}{^{52}C_2} = \dfrac{8}{663}$
Case V: There are 4 queens and 2 red coloured kings	$\dfrac{4}{52} \times \dfrac{2}{51} + \dfrac{2}{52} \times \dfrac{4}{51} =\dfrac{4}{663}$	$\dfrac{^{4}C_1 \times ^{2}C_1}{^{52}C_2} = \dfrac{8}{663}$

Answer: (I)$\dfrac{25}{102} ;$    (II) $\dfrac{13}{102} ;$    (III) $\dfrac{4}{663} ;$    (IV) $\dfrac{8}{663} ;$    (V) $\dfrac{8}{663}$

In a game show there are 12 doors. 3 of the doors have a car behind them. If a participant opens a door with a car, he's declared a winner and the game stops there. Aladdin can open 3 doors one after the other. What is the probability that Aladdin walks away a winner?

($1$) $\dfrac{21}{55}$            ($2$) $\dfrac{24}{55}$            ($3$) $\dfrac{31}{55}$             ($4$) $\dfrac{34}{55}$          

Solution

A person can win in $3$ ways – win of the $1^{\mathrm{st}}$ try, $2^{\mathrm{nd}}$ try or $3^{\mathrm{rd}}$ try. Note that to win in the $2^{\mathrm{nd}}$ try, one must have failed in the $1^{\mathrm{st}}$ try. Likewise, to win the $3^{\mathrm{rd}}$ try, one must have failed in the first $2$ tries.

The following are probabilities of winning in the $1^{\mathrm{st}}$ try $= \dfrac{3}{12}$ , $2^{\mathrm{nd}}$ try $= \dfrac{9}{12} \times \dfrac{3}{11}$ and $3^{\mathrm{rd}}$ try $= \dfrac{9}{12} \times \dfrac{8}{11}\times \dfrac{3}{10}$

Probability of winning
$= \dfrac{3}{12} + \dfrac{9}{12}\times \dfrac{3}{11} + \dfrac{9}{12}\times \dfrac{8}{11}\times \dfrac{3}{10}=\dfrac{1}{4} \times \left(1+\dfrac{9}{11}+\dfrac{36}{55}\right)$

$=\dfrac{1}{4}\times \dfrac{136}{55}= \dfrac{34}{55}$

Alternatively (Recommended Method)

There is only 1 way to lose, i.e., all 3 being the wrong doors.

Probability of Losing $= \dfrac{9}{12} \times \dfrac{8}{11}\times \dfrac{7}{10}=\dfrac{21}{55}$

Probability of Winning $=1- \dfrac{21}{55}= \dfrac{34}{55}$

Answer:($4$) $\dfrac{34}{55}$

Panam Airways' flights seat 300 passengers in 50 rows of 6 passengers each. What is the probability of 2 particular people sitting on the same row assuming all the seats are filled?

$(1$) $\dfrac{1}{6}$             ($2$) $\dfrac{5}{299}$            ($3$) $\dfrac{6}{299}$            ($4$) $\dfrac{10}{299}$          

Solution

Number of ways two seats can be selected in the flight $=$ $^{100}C_{2} = 150 \times 299$

Number of ways two seats can be selected in a particular row $=$ $^{6}C_{2} = 15$

Number of ways two seats in a particular row can be selected from 50 rows =$$50$\times$15$$=$750$

Therefore, P(two seats in the same row) =$\dfrac{750}{150 \times 299}=\dfrac{5}{299}$

Alternatively (Recommended Method)

If one passenger is already seated, the other passenger sits in one of the 5 seats (in the same row as the passenger) out of the remaining 299 seats. Therefore, the probability is$\dfrac{5}{299}$

Answer: ($3$) $\dfrac{5}{299}$

Class A has 4 boys and 6 girls. Class B has 3 boys and 5 girls. 2 students are randomly transferred from Class A to Class B. After this, 2 students are randomly selected from Class B. What is the probability that the 2 selected students are boys?

Solution

The final probability for selection of 2 boys from class B is dependent on whether the students transferred earlier from class A to B were both boys, both girls or 1 boy 1 girl.

Transfer	P(Transfer)	Students in B after transfer	P (Selecting 2 boys)	Overall Probability
2 boys	$\dfrac{4}{10} \times \dfrac{3}{9} =\dfrac{4}{30}$	5 boys & 5 girls	$\dfrac{5}{10} \times \dfrac{4}{9} =\dfrac{10}{45}$	$\dfrac{4}{30} \times \dfrac{10}{45} =\dfrac{40}{1350}$
2 girls	$\dfrac{6}{10} \times \dfrac{5}{9} =\dfrac{10}{30}$	3 boys & 7 girls	$\dfrac{3}{10} \times \dfrac{2}{9} =\dfrac{3}{45}$	$\dfrac{10}{30} \times \dfrac{3}{45} =\dfrac{30}{1350}$
1 boys & 1 girl	$2 \times \dfrac{6}{10} \times \dfrac{4}{9} =\dfrac{16}{30}$	4 boys & 6 girls	$\dfrac{3}{10} \times \dfrac{2}{9} =\dfrac{3}{45}$	$\dfrac{16}{30} \times \dfrac{6}{45} =\dfrac{96}{1350}$

Overall Probability =$\dfrac{40+30+96}{1350}= \dfrac{166}{1350}= \dfrac{83}{675}$

Answer: $\dfrac{83}{675}$