Probability
Probability
Allen Star Batch
Grade 12
Question:
Two persons $A$ and $B$ have respectively $n+1$ and $n$ coins, which they toss simultaneously. Then probability $P$ that $A$ will have more heads then $B$ belongs:
$\frac{1}{4} 0) = \frac{39}{64}$
$P(X_2 > 0) = \frac{37}{64}$
$P\left(\frac{X_1=2}{X_2=1}\right) = \frac{1}{5}$
$P\left(\frac{X_1=2}{X_2=1}\right) = \frac{3}{64}$
Step-by-Step Solution
Key Concept: By symmetry, P(λ > μ) = P(λ' < μ') where λ, λ' are heads/tails for A and μ, μ' are heads/tails for B. Since A has one extra coin, the extra coin outcome determines whether λ > μ, leading to P(λ > μ) = 1/2.
Let $\lambda, \mu$ be heads by A and B, with $\lambda + \lambda' = n+1$ and $\mu + \mu' = n$. The event $\lambda > \mu$ is equivalent to $\lambda' \mu) = \frac{1}{2}$.
Correct Answer: 1,4