# Simple Probability Question

#### jp023

##### New Member
OK, so I am wondering about something and am not good with probability, but maybe someone here has the answer

You have a die with only 4 sides, with the numbers 1-4 written on the sides. You roll the die 5 times. What is the probability of the number 2 showing up:

Once?
Twice?
Three times?
Four times?
All five times?

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#### Con-Tester

##### Member
Check out the Binomial distribution.

The probability of a success on one throw is P(Throw a 2) = 1/4, assuming the die is fair.

P(one 2 in 5 throws) = 5C1 × (1/4)^1 × (3/4)^4 = 405/1024 (≈ 0.396)
P(two 2’s in 5 throws) = 5C2 × (1/4)^2 × (3/4)^3 = 270/1024 (≈ 0.264)
P(three 2’s in 5 throws) = 5C3 × (1/4)^3 × (3/4)^2 = 90/1024 (≈ 0.088)
P(four 2’s in 5 throws) = 5C4 × (1/4)^4 × (3/4)^1 = 15/1024 (≈ 0.015)
P(five 2’s in 5 throws) = 5C5 × (1/4)^5 × (3/4)^0 = 1/1024 (≈ 0.001)

Can you see the pattern?

Note that the sum of the above probabilities is 781/1024 (≈ 0.763), which is the probability of throwing at least one 2 in five throws. If we apply the same formula to determine what the probability is of getting zero 2’s in five throws, we obtain:

P(zero 2’s in 5 throws) = 5C0 × (1/4)^0 × (3/4)^5 = 243/1024 (≈ 0.237)

Now 781/104 + 243/1024 = 1024/1024 = 1. This is not a coincidence.

#### jp023

##### New Member
you rock. thank you! 