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Probability question

2018 · 15 Apr · Shift 2 · Q32
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Probability question

2018 · 15 Apr · Shift 2 · Q32

JEE MainMathematicsProbabilityMCQ+4 / −1
A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the value of 'p' is :
  1. A
    15{1 \over 5}51​
  2. B
    13{1 \over 3}31​
  3. C
    25{2 \over 5}52​
  4. D
    14{1 \over 4}41​
View written solutionFree

Correct answer: B

  1. Let us compute the probability that XXX wins.

    • XXX starts.
    • XXX has probability ppp of getting a head on each turn.
    • YYY has a fair coin, so probability of head is 12\tfrac1221​.

    The game proceeds in rounds:

    • First, XXX tosses.
    • If XXX fails, then YYY tosses.
    • If both fail, the situation resets.

  2. Probability that both fail in one round

    • XXX gets tail with probability 1−p1-p1−p.
    • YYY gets tail with probability 12\tfrac1221​.

    So, probability both fail is

    (1−p)⋅12=1−p2.(1-p)\cdot \frac12 = \frac{1-p}{2}.(1−p)⋅21​=21−p​.

  3. Probability that XXX wins

    XXX can win:

    • immediately on the first toss: probability ppp
    • or after both fail once, then XXX wins: probability 1−p2⋅p\frac{1-p}{2}\cdot p21−p​⋅p
    • or after both fail twice, then XXX wins: probability (1−p2)2p\left(\frac{1-p}{2}\right)^2 p(21−p​)2p
    • and so on.

    Therefore,

    P(X)=p+p(1−p2)+p(1−p2)2+⋯P(X)=p+p\left(\frac{1-p}{2}\right)+p\left(\frac{1-p}{2}\right)^2+\cdotsP(X)=p+p(21−p​)+p(21−p​)2+⋯

    This is a geometric series with first term ppp and common ratio 1−p2\frac{1-p}{2}21−p​.

    Hence,

    P(X)=p1−1−p2.P(X)=\frac{p}{1-\frac{1-p}{2}}.P(X)=1−21−p​p​.

    Simplify the denominator:

    1−1−p2=2−(1−p)2=1+p2.1-\frac{1-p}{2}=\frac{2-(1-p)}{2}=\frac{1+p}{2}.1−21−p​=22−(1−p)​=21+p​.

    So,

    P(X)=p1+p2=2p1+p.P(X)=\frac{p}{\frac{1+p}{2}}=\frac{2p}{1+p}.P(X)=21+p​p​=1+p2p​.

  4. Given both players have equal winning probability

    Since one of them must eventually win,

    P(X)+P(Y)=1.P(X)+P(Y)=1.P(X)+P(Y)=1.

    If both probabilities are equal, then

    P(X)=P(Y)=12.P(X)=P(Y)=\frac12.P(X)=P(Y)=21​.

    Therefore,

    2p1+p=12.\frac{2p}{1+p}=\frac12.1+p2p​=21​.

  5. Solve for ppp

    Cross-multiplying,

    4p=1+p4p=1+p4p=1+p 3p=13p=13p=1 p=13.p=\frac13.p=31​.

  6. Check the options

    The correct option is

    13\boxed{\frac13}31​​

    which is Option B.

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