Equally Likely Events Examples - Word Problems

# Equally Likely Events Examples

Two events who has same theoretical probability are known as equally likely events. Suppose if we toss a coin, the events of getting a head or a tail are equally likely to happen. Equally likely events can be classified as independent events, mutually exhaustive events and mutually exclusive events. If A and B are two events having probability P(A) and P(B) then,

1) For independent events, P(A and B) = P(A)*P(B)

2) For mutually exclusive events, P(A or B) = P(A) + P(B)
P(A and B) = 0

## Word Problems

Problem 1:

Find the probability of getting a King when a card is chosen from a deck of 52 cards.

Solution:

Total number of events, n(S) = 52

Total number of likely events, n(E) = 4 [as there are four kings cards]

Probability of getting a king, P(E) = $\frac{n(E)}{n(S)}$ = $\frac{1}{13}$
Problem 2:

What is the probability of getting head when a coin is tossed and getting a 6 when a die is rolled?

Solution:

Probability of getting a head, P(A) = $\frac{1}{2}$

Probability of getting a six, P(B) = $\frac{1}{6}$

As the happening of one of these events do not affect the happening of the other event, these are independent events.

Probability of both the events happening together = P(A)*P(B) = $\frac{1}{2}$ $\times$ $\frac{1}{6}$ = $\frac{1}{12}$
Problem 3:

Find the probability of getting a king or a jack when one card is chosen from a deck of 52 cards.

Solution:

Probability of getting a king, P(A) = $\frac{4}{52}$ = $\frac{1}{13}$

Probability of getting a jack, P(B) = $\frac{4}{52}$ = $\frac{1}{13}$

As getting a king and getting jack are such events who cannot occur together.

Probability of either getting a king or a jack = P(A) + P(B) = $\frac{1}{13}$ + $\frac{1}{13}$ = $\frac{2}{13}$
Problem 4:

A bag has 3 red balls and 4 green balls. Find the probability of getting a red ball first and then a green ball.

Solution:

Probability of getting a red ball, P(A) = $\frac{3}{7}$

Probability of getting a red ball, P(B) = $\frac{4}{6}$ = $\frac{2}{3}$

As these events are independent of each other,

Probability of both events happening together = P(A)*P(B) = $\frac{3}{7}$ $\times$ $\frac{2}{3}$ = $\frac{2}{7}$