Probability Word Problems - Steps & Practice Problems

# Probability Word Problems

We will discuss different probability word problems here. The problems in probability that have to be deduced in mathematical form from the given statements in English are the probability word problems to be solved.

## Steps

How to solve a probability word problem:
We follow the following steps for solving a given word problem in probability.
1) The first step is to identify the experiment and the events for which we have to find the probability.
2) Now find out the number of outcomes of the experiment and every event that is determined in step 1.
3) Find the probability of all events determined in step 1 using values obtained in step 2.
4) Use probability laws to find the required probability as and when required like the multiplication rule, baye’s theorem, addition rule
etc.

## Practice Problems

Example 1: A box is filled with candies in different colors. We have 40 white candies, 24 green ones, 12 red ones, 24 yellow ones and 20 blue ones. If we have selected one candy from the box without peeking into it, find the probability of getting a green or red candy.

Solution: Let S be the sample space, A be the event of getting a green candy and B be the event of getting a red candy.

Now $n (S)$ = 40 + 24 + 12 + 24 + 20 = 120

Also, $n (A)$ = 24 and n (B) = 12

So $P (A)$ = $\frac{24}{120}$ = $\frac{1}{5}$

Also, $P (B)$ = $\frac{12}{120}$ = $\frac{1}{10}$

Clearly the two events are mutually exclusive so the probability of occurrence of either is the sum of both individually, that is,
P (A or B) = P (A) + P (B) = $\frac{1}{5}$ + $\frac{1}{10}$ = $\frac{3}{10}$.

Example 2: We have numbered cards from 1 to 20 and picked one at random. Find the probability that the card picked is numbered a multiple of 2 or 5.

Solution: Here, clearly n (S) = 20.

Let A be the event of drawing a card numbered multiple of 2

So,$A$ = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} => $n (A)$ = 10 => $P (A)$ = $\frac{10}{20}$

$B$ = event of drawing a card numbered multiple of 5

So, $B$ = {5, 10, 15, 20} => $n (B)$ = 4 => $P (B)$ = $\frac{4}{20}$

$A \cap B$ = {10, 20} => $n (A \cap B)$ = 2 => $P (A \cap B)$ = $\frac{2}{20}$

So $P (A or B)$ = $P (A) + P (B) - P (A \cap B)$

= $\frac{10}{20}$ + $\frac{4}{20}$ - $\frac{2}{20}$ = $\frac{12}{20}$ = $\frac{3}{5}$

Example 3: From a deck of 52 cards we draw 2 cards one by one without replacement. Find the probability that both cards are Aces.

Solution: Clearly, $n (S)$ = 52.

Let $A$ = event of drawing first king

=> $n (A)$ = 4 => $P (A)$ = $\frac{4}{52}$

Let $B$ = event of drawing second king when first is drawn

=> $n (B)$ = 3

Also we have drawn a card already so, $P (B)$ = $\frac{3}{51}$.

So the probability of occurrence of both A & B is given by the product of individual probabilities.

=> $P (A \cap B) = P (A) . P (B)$

= $\frac{4}{52}$ . $\frac{3}{51}$

= $\frac{1}{13}$ . $\frac{1}{17}$

= $\frac{1}{221}$