Probability is an important branch of mathematics which basically deals with the phenomena of chances or randomness of events. Probability of an event measures how likely it is expected to happen. An event is said to be "**most likely**" if it has higher probability and "**least likely**" if it has lower probability. The probability of an event varies between 0 and 1. An event with probability 0 is an **impossible** event, while event having probability 1 is said to be a **certain** event.

The formula for probability of an event**P(E)** is given below:

**Few sample problems based on probability theory are given below:**

**Problem 1:** A coin is tossed in air. What is the probability of getting 2 tails.

**Solution:** When a coin is tossed, the possible outcomes are:

Total number of outcomes = 4

We have,

$P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

Probability of getting 2 tails = $\frac{1}{4}$ = 0.25

**Problem 2:** A pair of dice is rolled. What is the probability of obtaining a sum of 6.

**Solution:** When a pair of dice is rolled, the total 36 outcomes are possible which are listed below:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Out of which (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) have total of 6.

We have,

$P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

Probability of getting a sum of 6 = $\frac{5}{36}$ = 0.14

**Problem 3:** A card is drawn from the deck of well shuffled cards at random. Find the probability of drawing an ace.

**Solution:** Total number of cards in a deck = 52

Total number of aces = 4

We have,

$P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

Probability of drawing an ace = $\frac{4}{52}$ = 0.08.

**Problem 4:** A bag contains 3 black, 5 red, 4 blue and 6 green marbles. One marble is drawn at random. What is the probability of NOT drawing a green ball.

**Solution:** Let us find the probability of drawing a green ball.

Total number of green balls = 6

Total balls = 3 + 5 + 4 + 6 = 18

We have,

$P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

Probability of drawing a green ball = $\frac{6}{18}$ = $\frac{1}{3}$

We know that $P(E)=1-P(Not\ E)$

Probability of NOT drawing a green ball = 1- $\frac{1}{3}$

= $\frac{2}{3}$

The formula for probability of an event

- Head - Head
- Head - Tail
- Tail - Head
- Tail - Tail

Total number of outcomes = 4

We have,

$P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

Probability of getting 2 tails = $\frac{1}{4}$ = 0.25

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Out of which (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) have total of 6.

We have,

$P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

Probability of getting a sum of 6 = $\frac{5}{36}$ = 0.14

Total number of aces = 4

We have,

$P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

Probability of drawing an ace = $\frac{4}{52}$ = 0.08.

Total number of green balls = 6

Total balls = 3 + 5 + 4 + 6 = 18

We have,

$P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

Probability of drawing a green ball = $\frac{6}{18}$ = $\frac{1}{3}$

We know that $P(E)=1-P(Not\ E)$

Probability of NOT drawing a green ball = 1- $\frac{1}{3}$

= $\frac{2}{3}$

Related Calculators | |

Calculation of Probability | Binomial Distribution Probability Calculator |

Binomial Probability Calculator | Coin Toss Probability Calculator |