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Probability question

2025 · 7 Apr · Shift 2 · Q30

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2025 · 7 Apr · Shift 2 · Q30

Probability question

2025 · 7 Apr · Shift 2 · Q30

JEE MainMathematicsProbabilityMCQ+4 / −1

A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is

\frac{m}{n}

,

\gcd(m, n) = 1

, then

n^2 - m^2

is equal to :

A
64
B
80
C
60
D
72

View written solutionFree

Correct answer: B

Define events

Let:

$U$ = the drawn coin is an unbiased coin
$D$ = the drawn coin is the double-headed coin
$H$ = head turns up on tossing

We need to find: $P(U\mid H)$

Prior probabilities

There are $19$ unbiased coins and $1$ double-headed coin, so total coins $=20$ .

Hence, $P(U)=\frac{19}{20}, \qquad P(D)=\frac{1}{20}$

Conditional probabilities of getting head

If the coin is unbiased, then $P(H\mid U)=\frac{1}{2}$
If the coin is double-headed, then $P(H\mid D)=1$

Total probability of head

Using total probability, $P(H)=P(H\mid U)P(U)+P(H\mid D)P(D)$

So, $P(H)=\frac{1}{2}\cdot \frac{19}{20}+1\cdot \frac{1}{20}$ $P(H)=\frac{19}{40}+\frac{2}{40}=\frac{21}{40}$

Apply Bayes' theorem

$P(U\mid H)=\frac{P(H\mid U)P(U)}{P(H)}$

Substituting, $P(U\mid H)=\frac{\frac{1}{2}\cdot \frac{19}{20}}{\frac{21}{40}}$ $P(U\mid H)=\frac{19}{40}\cdot \frac{40}{21}=\frac{19}{21}$

Thus, $\frac{m}{n}=\frac{19}{21}$ So, $m=19, \quad n=21$

Compute $n^2-m^2$

$n^2-m^2=21^2-19^2$ $=441-361=80$

Compare with options

$80$ corresponds to Option B.

Verification with stored answer

Stored correct answer is B, which matches our result.

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