Consider a coin P with probability P to be heads. What is the probability that…
Consider a coin P with probability P to be heads. What is the probability that the first head will appear on the even number of tosses ?
Answer: C. 1/(1-P)2 — Let q = 1 - P (the probability of a tail). The event "first head on an even-numbered toss" equals the union over k ≥ 1 of "first head at toss 2k." Probability…
- A.
1/(1-P)
- B.
P/(1-P)
- C.
1/(1-P)2
- D.
none of the above
Attempted by 2 students.
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Correct answer: C
Let q = 1 - P (the probability of a tail).
The event "first head on an even-numbered toss" equals the union over k ≥ 1 of "first head at toss 2k."
Probability first head at toss 2k = q^{2k-1} * P.
Sum over k: Σ_{k≥1} q^{2k-1} P = P*q * Σ_{k≥1} q^{2(k-1)} = P*q * Σ_{m≥0} (q^2)^m.
Use geometric series: P*q / (1 - q^2).
Simplify: 1 - q^2 = (1 - q)(1 + q) = P(1 + q), so the probability becomes q/(1 + q).
Final answer: (1 - P)/(2 - P).