Needs help on Russian Roullete Math Problem?

Basically, 2 persons are playing russian roulette, each are given 1 live bullet out of 6 chances. The rule is they must shoot at the same time.

1. What is the probability of both of them to die at the first round?

2. What is the probability of either of them die while the other survive at the first round ?

3. What is the probability of both of them survive the first round

Thank you in advance

2 Answers

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  • 6 months ago
    Favorite Answer

    Hello,

    In the first round, for each player, the probabilities are:

    - 1/6 to fire the bullet in the head;

    - 5/6 to survive.

    ► Thus the probability that both kill themselves on first round is:

    1/6 × 1/6 = 1/36 ≈ 2.78%

    ► Thus the probability that one of them survive and the other dies on first round is:

    1/6 × 5/6 + 5/6 × 1/6 = 10/36 ≈ 27.78%

    ► And the probability that both survive on first round is:

    5/6 × 5/6 = 25/36 ≈ 69.44%

    Regards,

    Dragon.Jade :-)

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  • Mike G
    Lv 7
    6 months ago

    1) P(Both die) = 1/6*1/6 = 1/36

    2) P(A dies B lives) = 1/6*5/6 = 5/36

    P(B dies A lives) = 1/6*5/6) = 5/36

    P = 10/36 = 5/18

    3) P = 5/6*5/6 = 25/36

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