The Mathematics of Maybe: Introduction to Probability | EOT

Question 12

Write the sample space and calculate the probability based on the given information.

(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?

(ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?

(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?

(iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?

(v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?

Solution

Understand the Question

The probability of an event $E$ is defined as:

$P(E) = \dfrac{\text{Number of favourable outcomes } n(E)}{\text{Total number of possible outcomes } n(S)}$

To solve each sub-part, identify the sample space $S$ , find the total number of outcomes $n(S)$ , list or count the favourable outcomes $n(E)$ that satisfy the given conditions, and then calculate $P(E) = \dfrac{n(E)}{n(S)}$ .

(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?

Step 1 · Find Total Outcomes and Favourable Outcomes

When two standard dice are rolled, the sample space consists of pairs $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$ : $n(S) = 6 \times 6 = 36$

The possible sums of two dice range from $2$ to $12$ . The prime numbers strictly greater than $5$ in this range are $7$ and $11$ .

List the favourable outcomes:

Sum $= 7$ : $\{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} \implies 6 \text{ outcomes}$
Sum $= 11$ : $\{(5, 6), (6, 5)\} \implies 2 \text{ outcomes}$

$n(E) = 6 + 2 = 8$

Step 2 · Calculate the Probability

\begin{aligned} P(\text{Sum is prime } > 5) &= \dfrac{n(E)}{n(S)} \\[0.6em] &= \dfrac{8}{36} \\[0.6em] &= \dfrac{2}{9} \end{aligned}

Answer

(i) $\dfrac{2}{9}$

(ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?

Step 1 · Find Total Outcomes

Total number of balls $= 4 + 3 + 2 = 9$ .

Number of ways to draw $2$ balls from $9$ balls without replacement:

\begin{aligned} n(S) &= \binom{9}{2} \\[0.6em] &= \dfrac{9 \times 8}{2 \times 1} = 36 \end{aligned}

Step 2 · Find Favourable Outcomes for Different Colours

Pairs of different colours can be formed in three ways:

1 Red and 1 Green: $\binom{4}{1} \times \binom{3}{1} = 4 \times 3 = 12$
1 Red and 1 Blue: $\binom{4}{1} \times \binom{2}{1} = 4 \times 2 = 8$
1 Green and 1 Blue: $\binom{3}{1} \times \binom{2}{1} = 3 \times 2 = 6$

Total favourable outcomes: $n(E) = 12 + 8 + 6 = 26$

Step 3 · Calculate the Probability

\begin{aligned} P(\text{Different colours}) &= \dfrac{n(E)}{n(S)} \\[0.6em] &= \dfrac{26}{36} \\[0.6em] &= \dfrac{13}{18} \end{aligned}

Answer

(ii) $\dfrac{13}{18}$

(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?

Step 1 · Write the Sample Space

When three coins are tossed, the sample space is: $S = \{\text{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}\}$ $n(S) = 2^3 = 8$

Step 2 · Identify Favourable Outcomes and Calculate Probability

Let $E$ be the event that the first coin shows heads and exactly two heads occur in total.

Outcomes with exactly two heads in total: $\{\text{HHT, HTH, THH}\}$

Among these, the outcomes where the first toss is Heads $(\text{H})$ are: $E = \{\text{HHT, HTH}\}$ $n(E) = 2$

\begin{aligned} P(E) &= \dfrac{n(E)}{n(S)} \\[0.6em] &= \dfrac{2}{8} \\[0.6em] &= \dfrac{1}{4} \end{aligned}

Answer

(iii) $\dfrac{1}{4}$

(iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?

Step 1 · Find Total and Favourable Number of Arrangements

Total 4-digit numbers formed using digits $\{1, 2, 3, 4\}$ without repetition: $n(S) = 4! = 4 \times 3 \times 2 \times 1 = 24$

For the number to be even, the units place must end in an even digit ( $2$ or $4$ ):

Units place: $2$ choices ( $2$ or $4$ )
Remaining $3$ places: Can be arranged in $3! = 6$ ways

Total favourable outcomes: $n(E) = 2 \times 3! = 2 \times 6 = 12$

Step 2 · Calculate the Probability

\begin{aligned} P(\text{Even number}) &= \dfrac{n(E)}{n(S)} \\[0.6em] &= \dfrac{12}{24} \\[0.6em] &= \dfrac{1}{2} \end{aligned}

Answer

(iv) $\dfrac{1}{2}$

Step 1 · Define Probabilities of Single Outcomes

For each question with $4$ options and $1$ correct option:

Probability of a correct answer, $p = \dfrac{1}{4}$
Probability of an incorrect answer, $q = 1 - \dfrac{1}{4} = \dfrac{3}{4}$

Step 2 · Calculate the Probability Using Binomial Probability

For $n = 3$ questions, the probability of getting exactly $k = 2$ correct answers is given by:

P(X = 2) = \binom{n}{k} p^k q^{n-k}

Substitute the values:

\begin{aligned} P(X = 2) &= \binom{3}{2} \left(\dfrac{1}{4}\right)^2 \left(\dfrac{3}{4}\right)^{3-2} \\[0.6em] &= 3 \times \dfrac{1}{16} \times \dfrac{3}{4} \\[0.6em] &= \dfrac{9}{64} \end{aligned}

Answer

(v) $\dfrac{9}{64}$

Common Mistakes

Prime Number Misconception: In part (i), forgetting that $2, 3, 5$ are primes $\le 5$ , so only sums of $7$ and $11$ should be counted.
Drawing Without Replacement: In part (ii), treating selections as independent with replacement (i.e. using denominator $9 \times 9 = 81$ instead of $\binom{9}{2} = 36$ ).
Order Restrictions: In part (iii), missing the condition that the first coin must be heads, leading to including $\text{THH}$ by mistake.

Question 12

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