Algebra

Probability

Medium

core

An urn contains 5 red and 5 black balls. A ball is drawn at random, its color is noted and is returned to the urn. Moreover, 2 additional balls of the same color are put in the urn and then a ball is drawn at random. The probability that the second drawn ball is red, is:

$\frac{5}{12}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

✅ Correct Option: 2

The urn initially contains 5 red balls and 5 black balls, for a total of 10 balls.

A ball is drawn at random, its color is noted, and it is returned to the urn along with 2 additional balls of the same color. Then a second ball is drawn.

The probability that the second ball is red depends on the color of the first ball drawn.

If the first ball drawn is red:

Probability of drawing red first: $P(\text{1st red}) = \frac{5}{10} = \frac{1}{2}$

After returning the red ball and adding 2 more red balls, the urn contains:

7 red balls
5 black balls
Total: 12 balls

Probability the second ball is red given the first was red:

$P(\text{2nd red | 1st red}) = \frac{7}{12}$

If the first ball drawn is black:

Probability of drawing black first: $P(\text{1st black}) = \frac{5}{10} = \frac{1}{2}$

After returning the black ball and adding 2 more black balls, the urn contains:

5 red balls
7 black balls
Total: 12 balls

Probability the second ball is red given the first was black:

$P(\text{2nd red | 1st black}) = \frac{5}{12}$

Using the law of total probability:

$P(\text{2nd red}) = P(\text{1st red}) \times P(\text{2nd red | 1st red}) + P(\text{1st black}) \times P(\text{2nd red | 1st black})$

$P(\text{2nd red}) = \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12}$

$P(\text{2nd red}) = \frac{7}{24} + \frac{5}{24}$

$P(\text{2nd red}) = \frac{12}{24}$

$P(\text{2nd red}) = \frac{1}{2}$

Therefore, the probability that the second drawn ball is red is $\frac{1}{2}$ .

2025: 21 May Shift 2

The probability that it will rain on any particular day is 50%. The probability that it rains only on the first 4 days of the week is:

2025: 21 May Shift 1

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be heart then the probability of the missing card to be a heart is:

2022: 17 Aug Shift 2

In a hospital, there are 300 patients, out of which 120 are female. It is known that out of 120 females, 10% of the patients are below 40 years of age. What is the probability that a patient chosen randomly is below 40 yrs of age given that the chosen patient is a female.

2025: 14 May Shift 1

If A is any event associated with sample space and If $E_1, E_2, E_3$ are mutually exclusive and exhaustive events. Then which of the following are true?

(A) $P(A) = P(E_1)P(E_2|A) + P(E_2)P(E_2|A) + P(E_3)P(E_3|A)$

(B) $P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)$

(D) $P(A|E_i) = \frac{P(E_i|A)P(E_i)}{\sum_{i=1}^3 P(E_i|A)P(E_i)}$ , $i = 1,2,3$

Choose the correct answer from the options given below:

2026: 22 May Shift 1

Bag I contains 2 red and 3 blue balls and bag II contains $\alpha$ red and 5 blue balls. One ball is drawn at random from one of the bags and is found to be blue. If the probability that it was drawn from bag II is $\frac{25}{52}$ , then $\alpha$ is equal to:

2026: 18 May Shift 2

One ticket is drawn from a purse containing four tickets with numbers 223, 133, 212, 322. Let $A_i$ be the event that i $^{th}$ digit of the number on drawn ticket is 2. Then which of the following events are independent ?

A. $A_1$ and $A_2$

B. $A_2$ and $A_3$

C. $A_1$ and $A_3$

D. $A_1$ , $A_2$ and $A_3$

Choose the correct answer from the options given below:

An urn contains 5 red and 5 black balls. A ball is drawn at random, its color is noted and is returned to the urn. Moreover, 2 additional balls of the same color are put in the urn and then a ball is drawn at random. The probability that the second drawn ball is red, is:

Solution

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