 # Probability Problems

Probability problems are very important for JEE exams. Probability talks about the outcome of an experiment. When you toss a coin, the outcome will be either heads or tails. The probability of an outcome can be determined by dividing the number of times the outcome has occurred by the total number of events. This article explains the concept of probability and its numerical applications.

## What is Probability?

The chance of an event is called probability. Probability can be used to predict the event likeliness. It always lies between 0 and 1. Higher probability indicates the higher chance of occurrence of an event. The concept of probability can be used for the events with a large number of outcomes and not restricted to less number of outcomes.

## Formula to find the Probability of an Event

Probability denotes the occurrence or non-occurrence of an event. The following formula is used in calculating the probability of an event.

Probability of event A = P(A) = (Number of outcomes favourable) / (Total number of outcomes)

Mutually exclusive events:

Two or more events associated to a random experiment are mutually exclusive if the occurrence of one prevents the occurrence of the other.Two or more events associated with a random experiment are mutually exclusive if there is no elementary event favourable to all events. Thus if two events are mutually exclusive then P(A ⋂ B) = 0

Likewise if A, B, C are mutually exclusive events, then  P(A ⋂ B⋂ C) = 0.

Mutually exhaustive events:

Two or more events associated to a random experiment are mutually exhaustive if their union is the sample space. I.e events A1, A2 ..An are associated to a random experiment with sample space S are exhaustive if A1 ⋃ A2 ⋃… An = S. All elementary events associated with a random experiment form a system of mutually exclusive and exhaustive events.

## Important Points to remember

1.If A and B are two events associated with a random experiment, then

P(A ⋃ B) = P(A)+P(B)-P(A ⋂ B)

2.If A and B are mutually exclusive events, then P(A ⋂B) = 0.

∴P(A ⋃ B) = P(A)+P(B)

3.If A, B and C are three events associated with a random experiment, then

P(A ⋃ B ⋃ C)= P(A)+P(B)+P(C)- P(A⋂ B)- P(B⋂C)- P(A⋂C)+P(A⋂B⋂C)

If A, B and C are mutually exclusive events, then

P(A⋂ B)= P(B⋂C)= P(A⋂C) = P(A⋂B⋂C) = 0

∴P(A ⋃ B ⋃ C) = P(A)+P(B)+P(C)

4.Let A and B be the two events associated to a random experiment, then

$P(\bar{A}\cap B)= P(B)-P(A\cap B)$

$P(\bar{A}\cap B)$ is called the probability of occurrence of B only.

$P(A\cap \bar{B})= P(A)-P(A\cap B)$

$P(A\cap \bar{B})$ is called probability of occurrence of A only.

$P((A\cap \bar{B})\cup (\bar{A}\cap B))$ = P(A) +P(B)-2P(A⋂ B)

$P((A\cap \bar{B})\cup (\bar{A}\cap B))$  is called the probability of occurrence of exactly one of two events A and B.

5.For any two events A and B the probability that exactly one of A,B occurs is given by

P(A) +P(B) – 2P(A⋃B) = P(A⋃B) -P(A⋂B)

6.If there are 3 events, A,B and C then

P(Atleast two of A, B, C occur) = P(A⋂B) +P(B⋂C) +P(C⋂A)-2P(A⋂B⋂C)

P(Exactly two of A, B, C occur) = P(A⋂B) +P(B⋂C) +P(C⋂A)-3P(A⋂B⋂C)

P(Exactly one of A, B, C occur) = P(A)+P(B)+P(C)+P(A⋂B)-2P(A⋂B) -2P(B⋂C)-2P(A⋂C)+3P(A⋂B⋂C)

7.If A and B are independent events associated with a random experiment, then

P(A/B) = P(A) and P(B/A) = P(B)

If A and B are two events two events associated with a random experiment, then

P(A⋂B) = P(A) P(B/A) , if P(A) ≠ 0

Or P(A⋂B) = P(B) P(A/B) , if P(B) ≠ 0.

8.If A and B are independent events then P(A/B) = P(A) and P(B/A) = P(B)

∴ P(A⋂B) = P(A) P(B)

P(A⋃B) = 1-$P(\bar{A})P(\bar{B})$

Related article:

Bayes’ theorem introduction

## Probability Problems Solved Examples

Example 1: A coin is tossed and a die is rolled. What is the probability that the coin shows the head and the die shows 3?

Solution:

When a coin is tossed, the outcome is either head or a tail. Similarly, when a die is rolled, the outcomes will be 1, 2, 3, 4, 5, 6.

Hence the required probability =(1/2) (1/6)=1/12.

Example 2: 8 nails and 12 nuts are present in a box. Among them, half of the nails and half of the nuts are rusted. If an item is chosen at random, what is the probability that it is rusted or is a nail?

Solution:

Total rusted items = 4 + 6 = 10;

Unrusted nails = 4.

Required probability = 4 + 10/8 + 12 = 14/20 = 7/10.

Example 3: The probability of happening an event A in one trial is 0.5. What is the probability that event A happens at least once in three independent trials?

Solution:

Here P(A)=0.5 and P($\overline{A}$)=0.5

The probability that A does not happen at all =(0.5)3

Thus, required Probability =$1-(0.5)^3$=0.875

Example 4: A man and a woman appear in an interview for two vacancies for the same post. The probability of man’s selection is 1/3 and that of the woman’s selection is 1/2. What is the probability that neither of them will be selected?

Solution:

Let E1 be the event that man will be selected and E2 the event that woman will be selected.

Then P(E1)=1/3 so P$\overline{E_1}$=1−1/3=2/3 and

P(E2)=1/2

So P($\overline{E_2}$)=1-1/2 = 1/2

Clearly E1 and E2 are independent events.

So, P($\overline{E_1}$$\overline{E_2}$)=P($\overline{E_1}$)×P($\overline{E_2}$)=2/3×1/2=1/3.

Example 5: Three dice are thrown simultaneously. What is the probability of obtaining a total of 17 or 18?

Solution:

Three dice can be thrown in 6×6×6=216 ways.

A total of 17 can be obtained as (5,6,6),(6,5,6), (6,6,5).

A total of 18 can be obtained as (6,6,6).

Hence the required probability =4/216=1/54.

Example 6: In order to get at least once a head with probability ≥0.9, what is the number of times a coin needs to be tossed?

Solution:

Probability of getting at least one head in n tosses

=> 1 − (1/2)n ≥ 0.9

⇒(1/2)n ≤ 0.1

⇒ 2n ≥ 10

⇒ n ≥ 3

Hence the least value of n = 4.

Example 7: The three ships namely A, B, and C sail from India to Africa. If the ratio of the ships reaching safely is 2: 5, 3: 7 and 6: 11, then find the probability of all of them arriving safely.

Solution:

We have a ratio of the ships A, B and C for arriving safely are 2: 5, 3: 7 and 6: 11 respectively.

The probability of ship A for arriving safely 2 /  2 / 7

Similarly, for 3 /  3 / 10 and for C = 6 /  6 / 17

Probability of all the ships for arriving safely = [2 / 7] × [3 / 10] × [6 / 17] = [18 / 595]

Example 8:  If A and B are two events such that P (A) = 0.4 , P (A + B) = 0.7 and P (AB) = 0.2, then P (B) =

Solution:

Since we have P (A + B) = P (A) + P (B) − P (AB)

⇒ 0.7 = 0.4 + P (B) − 0.2

⇒ P (B) = 0.5.

Example 9: If A and B are two independent events such that $P,(A\cap B’)=\frac{3}{25} \text \ and \ P(A’\cap B)=\frac{8}{25},$ then find P(A).

Solution:

Since events are independent.

So,  $P(A\cap {B}’)=P(A)\times P({B}’)=\frac{3}{25} \\\Rightarrow P(A)\times \{1-2P(B)\}=\frac{3}{25}$ …..(i)

Similarly,

$P(B)\times \{1-P(A)\}=\frac{8}{25}$  …..(ii)

On solving (i) and (ii), we get

$P(A)=\frac{1}{5} \text \ and \ \frac{3}{5}.$

Example 10: Consider two events A and B to be independent. The probability of occurrence of both the events A and B is 1 / 6 and the probability that neither of them occurs is 1 / 3. Then find the probability of the two events.

Solution:

$P(A\cap B)=P(A).P(B)=\frac{1}{6} \\P(\bar{A}\cap \bar{B})=\frac{1}{3}=1-P(A\cup B) \\\Rightarrow \frac{1}{3}=1-[P(A)+P(B)]+\frac{1}{6}\\\Rightarrow P(A)+P(B)=\frac{5}{6}. \\P(A) \text \ and \ P(B) \text \ are \ \frac{1}{2} \ and \frac{1}{3}.$

Example 11: There are 3 bags which are known to contain 2 white and 3 black balls; 4 white and 1 black balls and 3 white and 7 black balls respectively. A ball is drawn at random from one of the bags and found to be a black ball. Then the probability that it was drawn from the bag containing the most black balls is

Solution:

Consider the following events :

A → Ball drawn is black;

E1→ Bag I is chosen;

E2→ Bag II is chosen and

E3→ Bag III is chosen.

Then

$P({{E}_{1}})=({{E}_{2}})=P({{E}_{3}})=\frac{1}{3},\,\,P\left( \frac{A}{{{E}_{1}}} \right)=\frac{3}{5}. \\P\left( \frac{A}{{{E}_{2}}} \right)=\frac{1}{5},\,\,P\left( \frac{A}{{{E}_{3}}} \right)=\frac{7}{10}$

Required probability = $=P\left( \frac{{{E}_{3}}}{A} \right) \\=\frac{P({{E}_{3}})P(A/{{E}_{3}})}{P({{E}_{1}})P(A/{{E}_{1}})+P({{E}_{2}})P(A/{{E}_{2}})+P({{E}_{3}})P(A/{{E}_{3}})}\\=\frac{7}{15}.$