A box contains 10 coins, 4 of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. If the probability that the coin drawn is fair can be expressed as rational \frac{a}{b}, where a and b are relatively prime, then find the value of (a + b).

Engineering

Mathematics

Basic Definitions Related to Probability

Addition Theorem on Probability

Conditional Probability

Question

A box contains 10 coins, 4 of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. If the probability that the coin drawn is fair can be expressed as rational $\frac{a}{b}$ , where a and b are relatively prime, then find the value of (a + b).

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

[JEE' 2002(mains), Modified N=10, m=4, $\frac{9 m}{m + 8 N}$ ]

A = event what we have = coins drawn tossed twice, found to fall as HT
B₁ = It is a fair coins,       P (B₁) = $\frac{4}{10}$ = $\frac{2}{5}$
B₂ = It is a biased coins,   P (B₂) = $(1 - \frac{2}{5})$ = $\frac{3}{5}$
P (A / B₁) = $\frac{1}{4}$ ,   P (A / B₂) = $\frac{2}{9}$
P (B₁ / A) = $\frac{P (B_{1}) \cdot P (A / B_{1})}{P (B_{1}) \cdot P (A / B_{1}) + P (B_{2}) \cdot P (A / B_{2})}$ = $\frac{\frac{2}{5} \cdot \frac{1}{4}}{\frac{2}{5} \cdot \frac{1}{4} + \frac{3}{5} \cdot \frac{2}{9}}$ = $\frac{\frac{1}{10}}{\frac{1}{10} + \frac{2}{15}}$ = $\frac{3}{3 + 4}$ = $\frac{3}{7}$ ≡ $\frac{a}{b}$
Hence   a + b = 10