Question 1

Two archers, Arjun and Vikram, shoot arrows at a target. The probability of Arjun hitting the target is 2/3, and the probability of Vikram hitting the target is 5/6. Calculate the following probabilities:

a) Both hit the target.

b) Neither hits the target.

c) At least one hits the target.

d) Exactly one hits the target.

e) The target is hit.

f) Arjun hits and Vikram misses.

g) Only one of them hits the target.

Answer:

Ask yourself, can they both hit the target together? Yes, so clearly not mutually exclusive. Does the probability of Vikram hitting affect the probability of Arjun hitting or vice versa? No, they are two great professional shooters; why would one affect the other? Lol. Clearly independent events.

Since the events are independent, we use the multiplication rule for independent events and the complement rule where needed.

– Probability Arjun hits: P(A) = 2/3

– Probability Vikram hits: P(B) = 5/6

– Probability Arjun misses: P(A’) = 1 – 2/3 = 1/3

– Probability Vikram misses: P(B’) = 1 – 5/6 = 1/6

a) Both hit:

In simple English: Arjun hits the target AND Vikram hits the target.

$$P(A\cap B)=P\left(A\right)\times P\left(B\right)=\frac{2}{3}\times \frac{5}{6}=\frac{10}{18}=\frac{5}{9}$$

b) Neither hits:

In simple language: Arjun misses the target AND Vikram misses the target.

$$P(A^{‘}\cap B^{‘})=P\left(A^{‘}\right)\times P\left(B^{‘}\right)=\frac{1}{3}\times \frac{1}{6}=\frac{1}{18}$$

c) At least one hits:

As we already studied in previous articles, whenever we encounter an “at least one” condition, we use total – none, that is:

$$P\left(\text{at least one}\right)=1–P\left(\text{neither hits}\right)=1–\frac{1}{18}=\frac{17}{18}$$

d) Exactly one hits:

In simple language: (Arjun hits the target AND Vikram misses the target) OR (Vikram hits the target AND Arjun misses the target).

$$P\left(\text{exactly one}\right)=P(A\cap B^{‘})+P(A^{‘}\cap B)$$ $$=\left(P\right(A)\times P(B^{‘}\left)\right)+\left(P\right(A^{‘})\times P(B\left)\right)$$ $$=(\frac{2}{3}\times \frac{1}{6})+(\frac{1}{3}\times \frac{5}{6})$$ $$=\frac{2}{18}+\frac{5}{18}=\frac{7}{18}$$

e) Target is hit:

Same as at least one hitting (since the target is hit if at least one archer hits):

$$P\left(\text{target hit}\right)=\frac{17}{18}$$

f) Arjun hits, Vikram misses:

In simple language: Arjun hits the target AND Vikram misses the target.

$$P(A\cap B^{‘})=P\left(A\right)\times P\left(B^{‘}\right)=\frac{2}{3}\times \frac{1}{6}=\frac{2}{18}=\frac{1}{9}$$

g) Only one hits:

Same as part d, since “only one hits” means exactly one hits:

$$P(A\cap B^{‘})+P(A^{‘}\cap B)=\frac{7}{18}$$

Question 2

A spinner is divided into two equal sections: red and blue. Show that the probability of landing on red or blue is 1, and confirm that these events are mutually exclusive and exhaustive.

Answer:

When spinning the spinner, it can land on either red or blue, but not both simultaneously. Thus, the events are mutually exclusive.

Probability of landing on red: P(R) = 1/2

Probability of landing on blue: P(B) = 1/2

Total outcomes = 2 (red or blue), and each is equally likely.

Therefore, the probability of landing on Red OR Blue is:

Sum of probabilities in case of mutually exclusive events:

$$P\left(\text{R or B}\right)=P\left(R\right)+P\left(B\right)=\frac{1}{2}+\frac{1}{2}=1$$

Exhaustive: The spinner must land on either red or blue, so these events cover all possible outcomes, and the sum of probabilities of total possible outcomes is also 1, making them exhaustive.

Thus, the probability of landing on red or blue is 1.

Question 3

a) A fair eight-sided die, numbered 1 to 8, is rolled. What is the probability of rolling a 6 or an odd number?

Answer:

Rolling a 6 or an odd number (1, 3, 5, 7) cannot overlap since 6 is even.

Total outcomes: 8 (numbers 1 to 8).

Odd numbers: 1, 3, 5, 7 (4 outcomes).

Probability of rolling an odd number: P(odd) = 4/8 = 1/2

Probability of rolling a 6: 1/8

In case of mutually exclusive events, P(A or B) = P(A) + P(B):

$$P\left(\text{6 or odd}\right)=\frac{1}{8}+\frac{4}{8}=\frac{5}{8}$$

b) A fair eight-sided die, numbered 1 to 8, is rolled. What is the probability of rolling a 6 and an odd number?

Answer:

As we know, the events mentioned in the question are mutually exclusive. Focus on the keyword “and”; in case of mutually exclusive events, P(A ∩ B) = 0.

Also, by simple logic, we cannot get both 6 and an odd number together on a single throw, so the probability is 0.

Question 4

The probability that Priya forgets her math homework is 0.2. The probability that Sanjay forgets his math homework is 0.4. Calculate the probability that both Priya and Sanjay forget their homework on the same day.

Answer:

Can they forget their homework together? Yes, so clearly not mutually exclusive. Also, whether Priya forgets her homework does not affect whether Sanjay forgets his, so the events are independent.

P(Priya forgets) = 0.2

P(Sanjay forgets) = 0.4

In case of an “and” condition for independent events, P(A ∩ B) = P(A) × P(B):

$$P\left(\text{both forget}\right)=P\left(\text{Priya forgets}\right)\times P\left(\text{Sanjay forgets}\right)=0.2\times 0.4=0.08$$

Thus, the probability that both forget their homework is 0.08.

Question 5

Aisha has two jars of candies. She picks one candy from each jar. The probability of picking a chocolate from the first jar is 1/5, and the probability of picking a chocolate from both jars is 1/10. What is the probability of picking a chocolate from the second jar?

Answer:

The type of candy picked from one jar does not affect the other, so the events are independent.

P(chocolate from first jar) = 1/5

P(chocolate from both jars) = 1/10

Let P(chocolate from second jar) = P(C).

P(both) = P(first) × P(second):

$$\frac{1}{10}=\frac{1}{5}\times P\left(C\right)$$

$$P\left(C\right)=\frac{1}{10}÷\frac{1}{5}=\frac{1}{10}\times \frac{5}{1}=\frac{5}{10}=\frac{1}{2}$$

Thus, the probability of picking a chocolate from the second jar is 1/2.

Question 6

Events X and Y are defined as follows:

Event X: Rolling a die and landing on a number divisible by 3 (3 or 6).

Event Y: Drawing a heart from a standard deck of 52 cards.

Meera rolls a die and draws a card. Find the probability of both events X and Y occurring.

Answer:

Rolling a die and drawing a card are independent, as one does not affect the other.

Probability of Event X:

– Total outcomes on a six-sided die = 6.

– Numbers divisible by 3: 3, 6 (2 outcomes).

– P(X) = 2/6 = 1/3

Probability of Event Y:

– Total cards = 52, hearts = 13.

– P(Y) = 13/52 = 1/4

Probability of both:

$$P(X\cap Y)=P\left(X\right)\times P\left(Y\right)=\frac{1}{3}\times \frac{1}{4}=\frac{1}{12}$$

Thus, the probability of both events occurring is 1/12.

Question 7

The probability that a bus is late on any given day is 0.25. Find the probability that the bus is late for four consecutive days.

Answer:

The bus being late on one day does not affect other days, so the events are independent.

Probability of being late on one day: P(late) = 0.25

Probability of being late four days in a row:

$$P\left(\text{late on day 1 ∩ day 2 ∩ day 3 ∩ day 4}\right)=0.25\times 0.25\times 0.25\times 0.25=0.003906$$

Thus, the probability is 0.003906.

Question 8

The letters of the word “STATISTICS” are written on identical cards, shuffled, and one is selected at random. Calculate the probability of selecting an S or a vowel.

Answer:

Selecting one card means it cannot be both an S and a vowel (A, E, I, O, U) simultaneously, so the events are mutually exclusive.

Total letters: STATISTICS has 10 letters (S, T, A, T, I, S, T, I, C, S).

Probability of selecting an S: 3 S’s out of 10, so P(S) = 3/10

Probability of selecting a vowel: Vowels are A, I, I (3 vowels in the word STATISTICS), so P(vowel) = 3/10

Probability of S or vowel:

$$P\left(\text{S or vowel}\right)=P\left(S\right)+P\left(\text{vowel}\right)=\frac{3}{10}+\frac{3}{10}=\frac{6}{10}=\frac{3}{5}$$

Thus, the probability is 3/5.

Question 9

Ravi selects two students from a class of 20 girls and 30 boys. What is the probability that both students chosen are girls?

Answer:

Choosing the first student affects the total number of students and girls for the second choice, so the events are dependent.

Total students: 20 girls + 30 boys = 50 students.

Task: Ravi selects two students, and we want the probability that both are girls.

Key question: Why are the events of selecting the first girl and the second girl dependent?

Understanding Dependent Events

Two events are dependent if the outcome of the first event affects the probability of the second event. In contrast, events are independent if the outcome of one does not influence the other (e.g., flipping a coin twice, where the first flip doesn’t affect the second).

In this problem, the events are:

Event A: The first student selected is a girl.

Event B: The second student selected is a girl.

Why Are These Events Dependent?

The dependency arises because the selection of the first student changes the total number of students and the number of girls available for the second selection. Let’s break it down:

Total students = 50.

Number of girls = 20.

Probability of selecting a girl first (Event A):

$$P\left(\text{First is a girl}\right)=\frac{20}{50}$$

After Selecting the First Student:

Assume the first student selected is a girl (since we’re calculating the probability that both are girls).

After selecting one girl, the number of girls left is 20 – 1 = 19.

The total number of students left is 50 – 1 = 49.

Now, the probability of selecting a second girl (Event B, given Event A has occurred) is:

$$P\left(\text{Second is a girl | First is a girl}\right)=\frac{19}{49}$$

Dependency Explained:

The probability of the second student being a girl (19/49) is different from the probability of the first student being a girl (20/50). This change occurs because selecting the first girl reduces the number of girls and the total number of students available for the second selection.

Calculating the Probability

To find the probability that both students are girls, we multiply the probabilities of the dependent events:

$$P\left(\text{Both are girls}\right)=P\left(\text{First is a girl}\right)\times P\left(\text{Second is a girl | First is a girl}\right)$$ $$P\left(\text{Both are girls}\right)=\frac{20}{50}\times \frac{19}{49}=\frac{38}{245}$$

We will practice more questions on dependent events after we cover conditional probability.

Question 10

A weather station monitors two independent weather events: rain and strong winds. The probability of rain on a given day is 0.4, and the probability of strong winds is 0.3. Calculate the probability that on a given day:

a) Either it rains or there are strong winds (or both).

b) Neither event occurs.

Answer:

Since the events (rain and strong winds) are independent, we use the addition rule for the union of events and the multiplication rule for complements.

Probability of rain: P(R) = 0.4

Probability of strong winds: P(W) = 0.3

Probability of no rain: P(R’) = 1 – 0.4 = 0.6

Probability of no strong winds: P(W’) = 1 – 0.3 = 0.7

a) Either it rains or there are strong winds (or both):

For independent events, P(R ∪ W) = P(R) + P(W) – P(R ∩ W).

$$P(R\cap W)=P\left(R\right)\times P\left(W\right)=0.4\times 0.3=0.12$$

$$P(R\cup W)=0.4+0.3–0.12=0.58$$

Alternatively, P(R ∪ W) = 1 – P(neither) = 1 – (P(R’) × P(W’)) = 1 – (0.6 × 0.7) = 1 – 0.42 = 0.58

Thus, the probability is 0.58.

b) Neither event occurs:

$$P\left(\text{neither}\right)=P(R^{‘}\cap W^{‘})=P\left(R^{‘}\right)\times P\left(W^{‘}\right)=0.6\times 0.7=0.42$$

Thus, the probability is 0.42.