Amazing Solved Probability Questions with Answers

This post is about some solved probability questions. These questions make use of (i) the Addition Law of Probabilities, and (ii) the Multiplication Law of Probabilities.

Solved Probability Questions

Question 1: Box A contains 5 Green and 7 Red balls. Box B contains 3 Green, 3 Red, and 6 Yellow balls. A box is selected at random, and a ball is drawn at random from it. What is the probability that the bill drawn is green?

Solution:

Box A

Total Balls: 5 + 7 = 12
Prob(Green) = $\frac{3}{12}$

Box B

Total Balls: 3 + 3 + 6 = 12
P(Green) = $\frac{3}{12} = \frac{1}{4}$

$$P(A+B) = P(A) + P(B) = \frac{5}{12} + \frac{3}{12} = \frac{8}{12} = \frac{2}{3}$$

Question 2: A pair of fair dice is thrown twice. What is the probability of getting a total of 5 or 11?

Solution:

\begin{align*}
P(X = 11 \,\, or X = 5) &= P(X=11) + P(X=15) – P(X=11\,\,and\,\, X=5)\\
P(X=11) &= \frac{2}{36}\\
P(X=5) &= \frac{4}{36}=\frac{1}{9}\\
P(X=11\,\, and X=5) &= 0
\end{align*}

Therefore,

\begin{align*}
P(X=11\,\, or X=5) &= P(X=11) + P(X=5) \\
&=\frac{2}{36} + \frac{1}{9} = \frac{1}{6}
\end{align*}

Note that $P(X=11\,\, and X=5) = 0$, because the sum of two dice cannot be at the same time 5 and 11.

Question 3: A marble is drawn at random from a box containing 10 red, 30 white, 20 blue, and 15 orange marbles. What is the probability that it is (i) orange or red (ii) not red or blue (iii) not blue, (iv) white, (v) red, white, or blue.

Solution:

Total number of balls = 10 + 30 + 20 + 15 = 75
Number of Orange balls = 15
Number of Blue balls = 20
Number of White balls = 30
Number of Red balls = 10

P(a marble drawn is red or orange) = P(Red marble) + P(Orange marble)
$$=\frac{10}{75} + \frac{15}{75} = \frac{1}{3}$$
P(a marble drawn is not red or blue) = P(not Red) + P(Blue) – P(Blue and not Red)
$$=\frac{65}{75} + \frac{20}{75} – \frac{20}{75} = \frac{65}{75}$$
P(a ball drawn is not Blue) = $1 – P(Blue) = 1 – \frac{20}{75} = 0.733$
P(a ball drawn is white) = $\frac{30}{75}$
P(a ball drawn is Red, White, or Blue) = P(Red) + P(White) + P(Blue)
$$=\frac{10}{75} + \frac{30}{75} + \frac{20}{75} = \frac{60}{75}$$

Question 4: If two dice are thrown what are the various total number of dots that may turn up? What are the probabilities of each of them? What is the probability that the number of dots will total at least four?

Solution:

When two dice are thrown together, the minimum total number of dots is 2 (1, 1), and the maximum dots possible are 12 (6, 6). Therefore

Probability of 2 dots (1, 1) = $\frac{1}{36}$
Probability of 3 dots {(2, 1), (1, 2)} = $\frac{2}{36} = \frac{1}{18}$
Probability of 4 dots {(2,2) (3,1) (1,3)} = $\frac{3}{36} = \frac{1}{12}$
Probability of 5 dots {(4,1) (1,4) (2,3) (3,2)} = $\frac{4}{36} = \frac{1}{9}$
{Probability of 6 dots {(3,3) (4,2) (2,4) (5,1) (5,1)} = $\frac{5}{36}$
Probability of 7 dots {(4,3) (3,4) (5,2) (2,5) (6,1) (1,6)} = $\frac{6}{36} = \frac{1}{6}$
Probability of 8 dots {(6,2) (2,6) (5,3) (3,5) (4,4)} = $\frac{5}{36}$
Probability of 9 {(5,4) (4,5) (6,3) (3,6)} dots = $\frac{4}{36} = \frac{1}{9}$
Probability of 10 dots {(5,5) (6,4) (4,6)} = $\frac{3}{36} = \frac{1}{2}$
Probability of 11 dots {(5,6) (6,5)} = \frac{2}{36} = \frac{1}{18}$
Probability of 12 dots {(6,6)} = $\frac{1}{36}$
Probability that the number of dots will total at least 4 = $\frac{33}{36}$

Question 5: A one card is selected at random from a deck of 52 playing cards. What is the probability that the card is a club or a face card or both?

Solution:

\begin{align*}
P(club\,\, or\,\, face\,\, or\,\, both) &= P(club) + P(face) – P(club\,\, and\,\, face)\\
&=\frac{13}{52} + \frac{12}{52} – \frac{3}{52} = \frac{11}{26}
\end{align*}

Question 6: A class contains 10 men and 20 women of which half men and half women have brown eyes. What is the probability that a person chosen at random is a man or has brown eyes?

Solution:

Let $A$ be the event that it is a man (10 out of 30)
Let $B$ be the event that the person has brown eyes (5 men and 10 women: 15 out of 30)

$P(A\cap B)$ is a man AND has brown eyes $\frac{5}{30}$

\begin{align*}
P(A \cup B) &= P(A) + P(B) – P(A \cap B)\\
&= \frac{10}{30} + \frac{15}{30} – \frac{5}{30} = \frac{2}{3}
\end{align*}

Question 7: A drawer contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rested. If one item is chosen at random, what is the probability that it is rusted or is a bolt?

Solution:

Number of Bolts = 50
NUmber of Nuts = 150
Total number of Items = 50 + 150 = 200

Item chosen is rusted: $P(A) = \frac{100}{200} = \frac{1}{2}$
Item chosen is bolt: $P(B) = \frac{50}{200} = \frac{1}{4}$
Ite is Rusted and Bolt = $P(A\cap B) = P(A) \cdot P(B) = \frac{1}{2}\cdot \frac{1}{4} = \frac{1}{8}$

\begin{align*}
P(A \cup B) &= P(A) + P(B) – P(A\cap B) \\
&= \frac{1}{2} + \frac{1}{4} – \frac{1}{8} = \frac{5}{8}
\end{align*}