# You toss a die 6 times. What is the probability that you get no more than two 4’s?

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A die is tossed 6 times. The outcome of each toss lies in the set {1, 2, 3, 4, 5, 6}.

On each throw there are 6 outcomes with the probability of getting 4 equal to 1/6 and the probability of not getting a 4 equal to 5/6.

The desired result is that there should be no more than 2 fours. The conditions when this happens is if there are no 4s, if there is one 4 and if there are two 4's. The probability of this happening is `(5/6)^6` + `C(6, 1)*(1/6)*(5/6)^5` + `C(6, 2)*(1/6)^2*(5/6)^4` `~~ 0.9377`

**The probability that 4's are obtained no more than 2 times is approximately 0.9377.**