(7!/(4!*3!))*(1/2)^3 * (1/2)^4 = 35*(1/2)^7 which is about 0.2734375
When a die is rolled once, the probability of a 4 showing up is 1/6. Apply the binomial probability for finding the probability of exactly three fours out of 12 throws of a die. n=12 (number of throws) p=1/6 (probability of a four in a single throw) x = 3 (number of times out of 12 , a four showing up) P(x=3) = 12C3 (1/6)^3 (5/6)^(12-3) = 12C3 (1/6)^3 (5/6)^9 = 0.197443
If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.
John throws a fair 6-sided die. What is the probability he will get a multiple of 2?
It is 0.4
In the first two throws, the probability is (1/6)*(1/6) = 1/36. Eventually landing them in repeated throws is 1 (a theoretical certainty).
When a die is rolled once, the probability of a 4 showing up is 1/6. Apply the binomial probability for finding the probability of exactly three fours out of 12 throws of a die. n=12 (number of throws) p=1/6 (probability of a four in a single throw) x = 3 (number of times out of 12 , a four showing up) P(x=3) = 12C3 (1/6)^3 (5/6)^(12-3) = 12C3 (1/6)^3 (5/6)^9 = 0.197443
If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.
0.7969, 0.3355, 0.2936, 0.1468
If you throw a single fair coin multiple times, the probability of getting NO head is:For 1 throw: 1/2 For 2 throws: 1/2 squared = 1/4 For 3 throws: 1/2 cubed = 1/8 etc. The probability of getting AT LEAST ONE head is the complement; for example, for 3 throws, it would be 1 minus 1/8.
John throws a fair 6-sided die. What is the probability he will get a multiple of 2?
The probability is 0.0035
It is 0.4
In the first two throws, the probability is (1/6)*(1/6) = 1/36. Eventually landing them in repeated throws is 1 (a theoretical certainty).
Probability = (1/6)3 = 1/216 =0.00463
25%
1/2
1 in 64