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30 Sep 2003, 13:44
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Whats the probability that in three rolls of a fair die, 6 comes up at least twice.
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01 Oct 2003, 06:52
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Using Binomial Distribution,
The probability that an event will happen EXACTLY k times out of the n tries = nck * p^k * (1-p)^(n-k)
Probability of getting a 6 atleast 2 times = Probability of getting 6 exactly 2 times + Probability of getting 6 exactly 3 times
Probability of getting a 6 with a fair die = 1/6
Probability of getting 6 exactly 2 times = 3C2 (1/6)^2 + (1/6)^1 = 3/216
Probability of getting 6 exactly 3 times = 3C3 (1/6)^3 + (1/6)^0 = 1/216
So, Probability of getting a 6 atleast 2 times = 3/216 + 1/216 = 1/54
Am, I right?
The probability that an event will happen EXACTLY k times out of the n tries = nck * p^k * (1-p)^(n-k)
Probability of getting a 6 atleast 2 times = Probability of getting 6 exactly 2 times + Probability of getting 6 exactly 3 times
Probability of getting a 6 with a fair die = 1/6
Probability of getting 6 exactly 2 times = 3C2 (1/6)^2 + (1/6)^1 = 3/216
Probability of getting 6 exactly 3 times = 3C3 (1/6)^3 + (1/6)^0 = 1/216
So, Probability of getting a 6 atleast 2 times = 3/216 + 1/216 = 1/54
Am, I right?
01 Oct 2003, 07:58
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Probability of 6 coming atleast two times = P(2) + P(3)
where P(2), P(3) is probability of 6 coming 2 times and 3 times.
P(2) = 1/6*1/6*5/6 = 5/216 ----------- a
P(3) = 1/6*1/6*1/6 = 1/216 ------------b
Thus answer = a + b
= 1/36
where P(2), P(3) is probability of 6 coming 2 times and 3 times.
P(2) = 1/6*1/6*5/6 = 5/216 ----------- a
P(3) = 1/6*1/6*1/6 = 1/216 ------------b
Thus answer = a + b
= 1/36
