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The probability of getting a number of a die = 1/6
If at least 2 people should get the same number, it can happen in any of the following ways.
1) All four players getting the same number
2) Three players getting the same number
3) Two players getting the same number
Now let us calculate the probability of each way separately.
1) There is only one way of selecting 4 players out of 4 present.
Therefore probability that all 4 palyers get the same numer.
`P1 = 1 xx 1/6 xx 1/6 xx 1/6 xx 1/6`
`P1 = 1/1296`
2) There are 4C3 ways to select 3 players out of 4.
Therefore the probability that 3 players get the same number while other gets a different number.
`P2 = ^4C_3 xx 1/6 xx 1/6 xx 1/6 xx 5/6`
`P2 = 4 xx 5/1296`
`P2 = 20/1296`
3) There are 4C2 ways to select 2 players out of 4.
Therefore the probability that 2 players get the same number while other two get different numbers.
`P3 = ^4C_2 xx 1/6 xx 1/6 xx 5/6 xx 5/6`
`P3 = 12 xx 25/1296`
`P3 = 300/1296`
Therefore the probability that at least two players will get sme numbers is
`P = P1 + P2 + P3`
`P = 1/1296 + 20/1296 + 300/1296`
`P = 321/1296`
`P = 0.2477`
Therefore there is 24.77% chance that at least two players will get the same number.
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