A bag contains $(2n + 1)$ coins. It is known that $n$ of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is $\dfrac{31}{42}$, determine the value of $n$.

> [!TLDR] > The problem uses the law of total probability to combine the chances of picking a two-headed coin (always heads) and a fair coin (half heads). Solving the resulting equation gives $n = 10$. We have a bag with $2n+1$ coins. Of these, $n$ are two-headed (always show heads) and the remaining $(2n+1) - n = n+1$ are fair coins (one head, one tail). A coin is chosen at random and tossed once. The overall probability of getting a head is given as $\frac{31}{42}$. We need to find $n$. The key idea is **conditional probability** — the total probability of heads is the weighted average of the head probabilities from each type of coin, where the weights are the probabilities of picking that type. --- ### 1. Define the events Let: - $A$ = event that the chosen coin is two-headed. - $B$ = event that the chosen coin is fair. - $H$ = event that the toss shows a head. We know: - $P(A) = \frac{n}{2n+1}$ (since $n$ out of $2n+1$ coins are two-headed). - $P(B) = \frac{n+1}{2n+1}$ (the rest are fair). - $P(H \mid A) = 1$ (a two-headed coin always gives heads). - $P(H \mid B) = \frac{1}{2}$ (a fair coin gives heads half the time). --- ### 2. Apply the law of total probability The total probability of heads is: $$ P(H) = P(A) \cdot P(H \mid A) + P(B) \cdot P(H \mid B) $$ Substitute the values: $$ P(H) = \left( \frac{n}{2n+1} \right)(1) + \left( \frac{n+1}{2n+1} \right)\left( \frac{1}{2} \right) $$ --- ### 3. Simplify the expression Combine the terms over the common denominator $2n+1$: $$ P(H) = \frac{n}{2n+1} + \frac{n+1}{2(2n+1)} $$ Write the first term with denominator $2(2n+1)$: $$ P(H) = \frac{2n}{2(2n+1)} + \frac{n+1}{2(2n+1)} = \frac{2n + n + 1}{2(2n+1)} = \frac{3n + 1}{2(2n+1)} $$ --- ### 4. Set equal to the given probability …

A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides whereas the rest of

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(2n + 1)

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The problem uses the law of total probability to combine the chances of picking a two-headed coin (always heads) and a fair coin (half heads). Solving the resulting equation gives $n = 10$ .

We have a bag with $2n+1$ coins. Of these, $n$ are two-headed (always show heads) and the remaining $(2n+1) - n = n+1$ are fair coins (one head, one tail). A coin is chosen at random and tossed once. The overall probability of getting a head is given as $\frac{31}{42}$ . We need to find $n$ .

The key idea is conditional probability — the total probability of heads is the weighted average of the head probabilities from each type of coin, where the weights are the probabilities of picking that type.

1. Define the events

Let:

$A$ = event that the chosen coin is two-headed.
$B$ = event that the chosen coin is fair.
$H$ = event that the toss shows a head.

We know:

$P(A) = \frac{n}{2n+1}$ (since $n$ out of $2n+1$ coins are two-headed).
$P(B) = \frac{n+1}{2n+1}$ (the rest are fair).
$P(H \mid A) = 1$ (a two-headed coin always gives heads).
$P(H \mid B) = \frac{1}{2}$ (a fair coin gives heads half the time).

2. Apply the law of total probability

The total probability of heads is:

P(H) = P(A) \cdot P(H \mid A) + P(B) \cdot P(H \mid B)

Substitute the values:

P(H) = \left( \frac{n}{2n+1} \right)(1) + \left( \frac{n+1}{2n+1} \right)\left( \frac{1}{2} \right)

3. Simplify the expression

Combine the terms over the common denominator $2n+1$ :

P(H) = \frac{n}{2n+1} + \frac{n+1}{2(2n+1)}

Write the first term with denominator $2(2n+1)$ :

P(H) = \frac{2n}{2(2n+1)} + \frac{n+1}{2(2n+1)} = \frac{2n + n + 1}{2(2n+1)} = \frac{3n + 1}{2(2n+1)}

4. Set equal to the given probability …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 45 on this concept.

COMEDK 2025Set 2025-A1 markMCQ
$(n+1)$
›Reveal solutionSolution
The key idea is to treat the coin selection as a two‑case partition (two‑headed coin vs. fair coins) and apply the law of total probability. Solving the resulting equation gives $n = 5$ , so the correct option is (A).

Concept and intuition

We have a mixed bag: one trick coin that always lands heads, and $n$ fair coins that land heads with probability $\frac12$ . When we pick a coin at random and toss it, the overall chance of heads is a weighted average of the two cases. The weight for the trick coin is $\frac{1}{n+1}$ (since there are $n+1$ coins total), and for any fair coin it’s $\frac{n}{n+1}$ . The problem gives that overall probability as $\frac{7}{12}$ , so we set up an equation and solve for $n$ .

Step‑by‑step solution

Define the events Let $T$ be the event that the two‑headed coin is selected, and $F$ the event that a fair coin is selected. Since selection is random,

$P(T) = \frac{1}{n+1}, \qquad P(F) = \frac{n}{n+1}.$

Conditional probabilities for heads If the trick coin is chosen, heads is certain:

$P(\text{heads} \mid T) = 1.$
If a fair coin is chosen, the chance of heads is $\frac12$ :
$P(\text{heads} \mid F) = \frac12.$

Apply the law of total probability The overall probability of heads is

$P(\text{heads}) = P(T) \cdot P(\text{heads}\mid T) + P(F) \cdot P(\text{heads}\mid F).$
Substituting the values:
$P(\text{heads}) = \frac{1}{n+1} \cdot 1 + \frac{n}{n+1} \cdot \frac12.$

Simplify the expression

$P(\text{heads}) = \frac{1}{n+1} + \frac{n}{2(n+1)} = \frac{2 + n}{2(n+1)}.$

Set equal to the given probability The problem states this equals $\frac{7}{12}$ :
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COMEDK 2023Set 2023-E1 markMCQ
$\frac{23}{30}$
›Reveal solutionSolution

Computing $P(\text{white drawn}\mid H)=\frac45$ and $P(\text{white}\mid T)=\frac{11}{15}$ , Bayes' theorem gives $P(H\mid\text{white})=\frac{12}{23}$ .

Head (transfer 1 ball from A $(3W,2R)$ to B $(1W)$ , then draw from B's 2 balls):
$P(\text{white})=\tfrac35(1)+\tfrac25\left(\tfrac12\right)=\tfrac35+\tfrac15=\tfrac45.$
Tail (transfer 2 balls from A to B $(1W)$ , then draw from B's 3 balls):

$P(WW)=\tfrac{3}{10}\Rightarrow P(\text{white})=1,\quad P(WR)=\tfrac{6}{10}\Rightarrow \tfrac23,\quad P(RR)=\tfrac{1}{10}\Rightarrow \tfrac13.$ …
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COMEDK 2026Set 2026-A1 markMCQ
$\frac{21}{41}$
›Reveal solutionSolution

$P(E\mid A)=\tfrac35$ (without replacement), $P(E\mid B)=\tfrac{24}{49}$ (with replacement); Bayes gives $P(B\mid E)=\dfrac{40}{89}$ — option (C).

Likelihoods of drawing one yellow and one green ( $E$ ).

Jar A (3 yellow, 2 green; two draws without replacement):
$P(E\mid A)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=\frac{6}{10}=\frac{3}{5}.$
Jar B (4 yellow, 3 green; two draws with replacement):
$P(E\mid B)=2\cdot\frac{4}{7}\cdot\frac{3}{7}=\frac{24}{49}.$
Bayes' theorem with $P(A)=P(B)=\tfrac12$ (the $\tfrac12$ cancels): …
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COMEDK 2024Set 2024-E1 markMCQ
$\frac{1}{2}$
›Reveal solutionSolution
Given the first toss is a tail, the coin is tossed a third time only if the second toss is also a tail — probability $\tfrac12$ .

The coin stops the moment a head appears, or after three tosses. We are told the first toss is a tail (no head on toss 1).

Now the second toss decides:

Head on toss 2 $\Rightarrow$ stop after 2 tosses.

Tail on toss 2 $\Rightarrow$ a third toss is made. …
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COMEDK 2024Set 2024-M1 markMCQ
$\frac{17}{30}$
›Reveal solutionSolution
Conditioning on the first two draws (white -> not replaced; black -> replaced plus one extra black) and using total probability gives $P(\text{3rd black}) = \tfrac{23}{30}$ — option (D).

Rules. Start with 2 white, 2 black (total 4). Drawing white removes it (whites $-1$ , total $-1$ ). Drawing black puts it back and adds one more black (blacks $+1$ , total $+1$ ).

First two draws and the composition just before the 3rd draw

WW: $\tfrac24\cdot\tfrac13 = \tfrac16$ ; urn becomes W0, B2 (total 2) $\Rightarrow P(\text{black}) = 1$ .

WB: $\tfrac24\cdot\tfrac23 = \tfrac13$ ; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$ .

BW: $\tfrac24\cdot\tfrac25 = \tfrac15$ ; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$ .

BB: $\tfrac24\cdot\tfrac35 = \tfrac{3}{10}$ ; urn becomes W2, B4 (total 6) $\Rightarrow P(\text{black}) = \tfrac23$ . …
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COMEDK 2024Set 2024-E1 markMCQ
$\frac{1}{6}$
›Reveal solutionSolution
This is a classic conditional probability problem (often called the "three cards" or "Bertrand's box" variant). The key is that seeing a red side updates the probability space to only the red sides, and among those, only one belongs to the mixed card. The answer is 1/3.

We are asked: given that the visible side is red, what is the probability that the other side is black? This is not simply "one of the two remaining cards has a black other side" because the cards are not equally likely once we condition on the observation.

Concept and Intuition

The pitfall is to think: "We see red, so the card is either the all-red or the mixed card. That's two possibilities, so the chance is 1/2." But this ignores that the all-red card has two red sides, while the mixed card has only one red side. When we pick a card at random and then look at a random side, each of the six sides is equally likely to be the one we see. Seeing red eliminates the three black sides, leaving only the three red sides. Among those three red sides, two belong to the all-red card and only one belongs to the mixed card. So the probability that the other side is black is 1/3.

Step-by-step reasoning

Label the sides.

Let the cards be:

Card A: both sides red (sides $R_1, R_2$ )

Card B: both sides black (sides $B_1, B_2$ )

Card C: one red, one black (sides $R_3, B_3$ )

Count equally likely outcomes.

When we pick a card uniformly at random and then look at a random side, there are $3 \times 2 = 6$ equally likely side-views. Each of the six sides has probability $1/6$ of being the one we see.

Condition on seeing a red side.

The red sides are: $R_1, R_2$ (from card A) and $R_3$ (from card C). That's 3 red sides. So the conditional space has 3 equally likely possibilities. …
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KCET 2019Set A-11 markMCQ
$S = \{1, 2, 3, 4, 5, 6, 7\}$
›Reveal solutionSolution
By Bayes' theorem the probability is $\frac{3}{5}$ — option (D).

The set $S=\{1,2,3,4,5,6,7\}$ has $3$ even numbers $(2,4,6)$ and $4$ odd numbers, and one number is picked at random.

Let $E$ = "the number is even" and $R$ = "the man reports it as even."

Priors.
$P(E)=\frac{3}{7},\qquad P(E^c)=\frac{4}{7}.$
Likelihoods. He tells the truth with probability $\frac{2}{3}$ and lies with probability $\frac{1}{3}$ .

If the number really is even, a truthful report says "even": $P(R\mid E)=\frac{2}{3}$ .

If the number is odd, only a lie reports "even": $P(R\mid E^c)=\frac{1}{3}$ .

Bayes' theorem. …
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KCET 2025Set A-11 markMCQ
$\dfrac{2}{5}$
›Reveal solutionSolution

The friend has already been met (the effect); we want the probability of the cause (temple B) — that reversal of conditioning is exactly Bayes' theorem.

Step 1 — Name the events.

Let $A$ = "Meera visits temple A", $B$ = "Meera visits temple B", $F$ = "she meets her friend".

She visits only one of the two temples, so $A$ and $B$ are mutually exclusive and exhaustive:
$P(A)=\frac25\quad\Longrightarrow\quad P(B)=1-\frac25=\frac35 .$
Given: $P(F\mid A)=\dfrac13$ , $P(F\mid B)=\dfrac27$ .

Step 2 — Why Bayes.

We are told the outcome ( $F$ happened) and asked for the probability of a cause ( $B$ ). Bayes' theorem inverts the conditioning:
$P(B\mid F)=\frac{P(B)\,P(F\mid B)}{P(A)\,P(F\mid A)+P(B)\,P(F\mid B)} .$
The denominator is $P(F)$ by the law of total probability — the friend can be met on either branch.

Step 3 — Compute the two branch probabilities.
$P(A\cap F)=\frac25\times\frac13=\frac{2}{15},\qquad P(B\cap F)=\frac35\times\frac27=\frac{6}{35} .$
Step 4 — Total probability of meeting the friend. …
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COMEDK 2026Set 2026-M1 markMCQ
$\frac{\mathbf{1}}{\mathbf{3}}$
›Reveal solutionSolution
Summing the three disjoint "exactly one win" paths gives $\dfrac{1}{12}+\dfrac{1}{18}+\dfrac{1}{8}=\dfrac{19}{72}$ — option (D).

Set up the conditional probabilities. Let $W_i$ mean "defeats head $i$ " and $L_i$ mean "fails":

$P(W_1) = \dfrac{1}{3}$ , so $P(L_1) = \dfrac{2}{3}$ .

After a win: next-head win probability $=\dfrac{2}{3}$ , so next-head loss probability $=\dfrac{1}{3}$ .

After a loss: next-head win probability $=\dfrac{1}{4}$ , so next-head loss probability $=\dfrac{3}{4}$ .

Earning the medal means exactly one of the three heads is defeated. The three disjoint sequences are $W_1L_2L_3$ , $L_1W_2L_3$ , and $L_1L_2W_3$ .

Path 1 — $W_1 L_2 L_3$ (win, then loss after a win, then loss after a loss):
$P = \frac{1}{3}\cdot\frac{1}{3}\cdot\frac{3}{4} = \frac{3}{36} = \frac{1}{12}$
Path 2 — $L_1 W_2 L_3$ (loss, then win after a loss, then loss after a win): …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. An urn contains 5 red and 5 black coloured balls. A ball is picked at random, its colour noted and then is put back into the urn. Now, additional two balls of the same colour are put into the urn and then a ball is drawn at random. What is the probability that the ball now chosen is red. (A) 7/24 (B) 1/2 (C) 5/8 (D) 2/3$
›Reveal solutionSolution
By symmetry the second-draw red probability averages to $\tfrac12$ .

Start: 5 red, 5 black (10 total). First ball is drawn, replaced, then 2 more of the same colour are added, so the urn always has 12 balls before the second draw.

First ball red ( $P=\tfrac{5}{10}=\tfrac12$ ): urn now 7R, 5B ⇒ $P(\text{red})=\tfrac{7}{12}$ . …
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COMEDK 2025Set 2025-A1 markMCQ
$i$
›Reveal solutionSolution
This is a classic Bayes’ theorem problem: we are given prior probabilities for choosing each bag and the conditional probabilities of drawing a white ball from each bag; after observing a white ball, we want the posterior probability that it came from Bag III. The answer is $\frac{9}{11}$ .

We start by noting that the problem gives us three bags with different compositions, and a rule for choosing a bag: the probability of picking bag $i$ is $\frac{i}{6}$ . Then, from that bag, we pick a ball uniformly at random. We are told that a white ball was selected, and we need the probability that it came from Bag III. This is a textbook case for Bayes’ theorem, which lets us “reverse” conditional probabilities: we know $P(\text{white} \mid \text{bag } i)$ , and we want $P(\text{bag } i \mid \text{white})$ .

Let’s work through it step by step.

Define events and write down given probabilities. Let $B_i$ be the event that bag $i$ is chosen ( $i=1,2,3$ ). We are told:

$P(B_1) = \frac{1}{6},\quad P(B_2) = \frac{2}{6},\quad P(B_3) = \frac{3}{6}.$
Let $W$ be the event that a white ball is drawn.

From the bag compositions:

Bag I: 3 red, 0 white → $P(W \mid B_1) = 0$ .

Bag II: 2 red, 1 white → $P(W \mid B_2) = \frac{1}{3}$ .

Bag III: 0 red, 3 white → $P(W \mid B_3) = 1$ .

Compute the total probability of drawing a white ball. By the law of total probability:

$P(W) = P(B_1)P(W\mid B_1) + P(B_2)P(W\mid B_2) + P(B_3)P(W\mid B_3).$
Substitute:
$P(W) = \frac{1}{6}\cdot 0 + \frac{2}{6}\cdot\frac{1}{3} + \frac{3}{6}\cdot 1 = 0 + \frac{2}{18} + \frac{3}{6} = \frac{1}{9} + \frac{1}{2}.$
Get a common denominator (18):
$P(W) = \frac{2}{18} + \frac{9}{18} = \frac{11}{18}.$

Apply Bayes’ theorem to find $P(B_3 \mid W)$ . Bayes’ theorem says:

$P(B_3 \mid W) = \frac{P(B_3)\,P(W\mid B_3)}{P(W)}.$
Plug in the numbers:
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KCET 2021Set A-11 markMCQ
$\dfrac{56}{73}$
›Reveal solutionSolution
This is a reverse-probability question (effect → cause), so use Bayes' theorem with the total probability of a standard-quality car in the denominator.

Step 1 — Define the events.

$X$ : the car came from plant X — $P(X) = 0.70$

$Y$ : the car came from plant Y — $P(Y) = 0.30$

$S$ : the car is of standard quality

The conditional (likelihood) data given:
$P(S\mid X) = 0.80, \qquad P(S\mid Y) = 0.90$
Note $X$ and $Y$ are mutually exclusive and exhaustive ( $0.7 + 0.3 = 1$ ), which is exactly what Bayes' theorem needs.

Step 2 — Why Bayes and not simple conditioning.

We are told the effect (the chosen car is standard) and asked for the probability of the cause (it came from X). That inversion — $P(X\mid S)$ from $P(S\mid X)$ — is precisely Bayes' theorem:
$P(X \mid S) = \frac{P(X)\,P(S\mid X)}{P(X)\,P(S\mid X) + P(Y)\,P(S\mid Y)}$
Step 3 — Compute the numerator.
$P(X)P(S\mid X) = 0.70 \times 0.80 = 0.56$
Step 4 — Compute the denominator (total probability of a standard car).
$P(S) = 0.70\times0.80 + 0.30\times0.90 = 0.56 + 0.27 = 0.83$
Step 5 — Divide. …
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Q.A bag contains $(2n + 1)$ coins. It is known that $n$ of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is $\dfrac{31}{42}$ , determine the value of $n$ .

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

1. Define the events

2. Apply the law of total probability

3. Simplify the expression

4. Set equal to the given probability …

Method: Law of total probability to build an equation in an unknown

Steps

Common Mistakes

Concept and Intuition

Step-by-step reasoning

1. Define the events

2. Apply the law of total probability

3. Simplify the expression

4. Set equal to the given probability …

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