Suppose a Girl Throws a Die. If She Gets 5 or 6 She Tosses a Coin Three Times and Notes the Number of Heads; if She Gets 1, 2, 3 or 4 She Tosses a Coin Once. If She Obtained Exactly One Head, What Is the Probability That She Threw 1, 2, 3 or 4 With the Die

NCERT Class 12 Mathematics — Probability, Exercise 13.3.

Problem

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3, or 4 with the die?

Key insight. “Exactly one head” can happen two entirely different ways depending on which branch of the experiment was followed — one head out of three tosses, or one head out of a single toss — and each branch has its own, different probability of producing that outcome. Bayes’ theorem is built exactly for combining these into the probability of “which branch, given the outcome.”

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Approach

Let $D$ be the event “die shows 5 or 6” and $D’$ be “die shows 1, 2, 3, or 4,” with known prior probabilities from a fair die. Find the probability of “exactly one head” under each branch separately — a binomial probability for three tosses under $D$, and a simple $\tfrac12$ under $D’$. Combine these using Bayes’ theorem to find $P(D’\mid E)$, where $E$ is “exactly one head.”

Solution

Step 1 — Find the prior probabilities of each branch

$$P(D) = \frac{2}{6}=\frac13 \quad \text{(die shows 5 or 6)}$$

$$P(D’) = \frac{4}{6}=\frac23 \quad \text{(die shows 1, 2, 3, or 4)}$$

Step 2 — Find P(E | D): exactly one head in three coin tosses

$$P(E\mid D) = \binom{3}{1}\left(\frac12\right)^3 = 3\times\frac18 = \frac38$$

Step 3 — Find P(E | D’): exactly one head in a single coin toss

$$P(E\mid D’) = \frac12$$

Step 4 — Apply Bayes’ theorem

$$P(D’\mid E) = \frac{P(D’)\,P(E\mid D’)}{P(D)\,P(E\mid D)+P(D’)\,P(E\mid D’)}$$

Step 5 — Substitute the values

$$P(D’\mid E) = \frac{\dfrac23\times\dfrac12}{\dfrac13\times\dfrac38+\dfrac23\times\dfrac12}$$

Step 6 — Simplify the numerator and each term of the denominator

Numerator: $\dfrac23\times\dfrac12=\dfrac13$

Denominator, first term: $\dfrac13\times\dfrac38=\dfrac18$

Denominator, second term: $\dfrac23\times\dfrac12=\dfrac13$

Step 7 — Combine over a common denominator (24) and simplify

$$P(D’\mid E) = \frac{\dfrac13}{\dfrac18+\dfrac13} = \frac{\dfrac{8}{24}}{\dfrac{3}{24}+\dfrac{8}{24}} = \frac{8/24}{11/24} = \frac{8}{11}$$

Answer

Common mistakes

• Using $P(E\mid D)=\dfrac12$ for both branches. The probability of getting exactly one head is different depending on how many times the coin was tossed — one toss gives $\dfrac12$, but three tosses give $\binom31\left(\dfrac12\right)^3=\dfrac38$, a genuinely different value that must be computed with the binomial formula.
• Forgetting to normalise by the total probability of $E$ across both branches. $P(D’)\times P(E\mid D’)$ alone gives $P(D’\cap E)$, not $P(D’\mid E)$ — Bayes’ theorem requires dividing by the sum of this quantity across all branches that could produce $E$.

Practise next

• A bag contains a fair coin and a biased coin that always shows heads. A coin is picked at random and tossed twice. If both tosses show heads, find the probability the biased coin was picked, using the same Bayes’ theorem setup.