# CRE 6 - Conditional Probability | Modern Math - Probability

**CRE 6 - Conditional Probability | Modern Math - Probability**

2 unbiased die are rolled simultaneously. The numbers that appear on the two dice are different. Find the probability that the sum is 6.

- A.
5/36

- B.
1/9

- C.
2/15

- D.
None of these

Answer: Option C

**Explanation** :

When two die are rolled total possible outcomes = 6 × 6 = 36.

Out of these 36 outcomes, 6 outcomes result in same number on both dice.

Since the number on two dice are different, possible outcomes here = 36 – 6 = 30.

∴ Desired outcomes for sum to be 6 = {(1, 5) or (5, 1) or (2, 4) or (4, 2)} i.e., 4 outcomes.

∴ Required probability = 4/30 = 2/15

Hence, option (c).

Workspace:

**CRE 6 - Conditional Probability | Modern Math - Probability**

2 integers are selected at random from the integers 1 to 11. The sum of these 2 integers is even. Find the probability that both the selected integers are odd.

- A.
2/5

- B.
3/5

- C.
5/22

- D.
None of these

Answer: Option B

**Explanation** :

There are 11 integers: 6 odd & 5 even.

Sum of two integers is even when either both are even or both are odd.

**Both even**: Number of ways of selecting two even integers = ^{5}C_{2} = 10.

**Both odd**: Number of ways of selecting two odd integers = ^{6}C_{2} = 15.

∴ Total possible outcomes = 10 + 15 = 25.

∴ Desired outcomes for both integers to be odd = 15.

∴ Required probability = 15/25 = 3/5

Hence, option (b).

Workspace:

**CRE 6 - Conditional Probability | Modern Math - Probability**

An unbiased dice is rolled twice and the sum of the two roll outcomes is 6. What is the conditional probability that the number 4 has appeared at least once?

- A.
2/5

- B.
3/5

- C.
5/22

- D.
None of these

Answer: Option A

**Explanation** :

Total possible outcomes for sum to be 6 = {(1, 5) or (5, 1) or (2, 4) or (4, 2) or (3, 3)} i.e., 5 outcomes.

Here, desired outcomes = {(2, 4) or (4, 2)} i.e., 2 outcomes.

∴ Required probability = 2/5

Hence, option (a).

Workspace:

**CRE 6 - Conditional Probability | Modern Math - Probability**

2 cards are drawn at random from a well shuffled pack of 52 cards. Both the cards turn out to be red. Find the probability that both the cards have different numbers on them. (Assume that Ace, Jack, Queen & King are not considered as numbered cards.)

- A.
18/65

- B.
64/325

- C.
144/325

- D.
288/325

Answer: Option C

**Explanation** :

Total possible outcomes when both cards are red = ^{26}C_{2} = 325.

Now, there are 18 red number cards available.

∴ Desired outcomes for both the cards to have different numbers = ^{18}C_{2} – 9 = 153 – 9 = 144

[out of total ^{18}C_{2} ways of selecting number cards 9 ways will give us same number on both cards.]

∴ Required probability = 144/325

Hence, option (c).

Workspace:

**CRE 6 - Conditional Probability | Modern Math - Probability**

Three urns contain red and black balls. The first urn contains 6 red & 4 black balls. The second urn contains 4 red & 6 black balls. The third urn contains 5 red & 5 black balls. An urn is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the first urn.

- A.
2/5

- B.
4/5

- C.
4/15

- D.
1/3

Answer: Option A

**Explanation** :

Let us calculate P (selecting 1^{st} urn and drawing a red ball from it).

Out of three urns probability of selecting 1^{st} urn = 1/3.

P (drawing a red ball from 1^{st} urn) = $\frac{6}{10}$ = $\frac{3}{5}$

P (selecting 1^{st} urn and drawing a red ball from it) = $\frac{1}{3}\times \frac{3}{5}$ = $\frac{3}{15}$

Similarly,

P (selecting 2^{nd} urn and drawing a red ball from it) = $\frac{1}{3}\times \frac{2}{5}$ = $\frac{2}{15}$

P (selecting 3^{rd} urn and drawing a red ball from it) = $\frac{1}{3}\times \frac{1}{2}$ = $\frac{1}{6}$

P (If the ball drawn is red, find the probability that it is drawn from the first urn)

= $\frac{\mathrm{P}\left(\mathrm{red}\mathrm{}\mathrm{ball}\mathrm{}\mathrm{from}\mathrm{}1\mathrm{st}\mathrm{}\mathrm{urn}\right)}{\mathrm{P}\left(\mathrm{red}\mathrm{}\mathrm{ball}\mathrm{}\mathrm{from}\mathrm{}1\mathrm{st}\mathrm{}\mathrm{urn}\right)+\mathrm{P}\left(\mathrm{red}\mathrm{}\mathrm{ball}\mathrm{}\mathrm{from}\mathrm{}2\mathrm{nd}\mathrm{}\mathrm{urn}\right)+\mathrm{P}\left(\mathrm{red}\mathrm{}\mathrm{ball}\mathrm{}\mathrm{from}\mathrm{}3\mathrm{rd}\mathrm{}\mathrm{urn}\right)}$

= $\frac{\frac{3}{15}}{\frac{3}{15}+\frac{2}{15}+\frac{1}{6}}$ = $\frac{\frac{3}{15}}{\frac{6+4+5}{30}}$ = $\frac{3}{15}\times \frac{30}{15}$ = $\frac{2}{5}$

Hence, option (a).

Workspace:

## Feedback

**Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.**