Ok so i have been taught this formula regarding binomial probability
Repeat an even n times
x = # of successes p = probability of successes q = probability of failure
p(x = a) = nCa * p^a * q^(n-a)
Ok wow seems good. So i try to solve this question using this formula.
A pair of fair dice is rolled 10 times. Let X be the number of rolls in which we see at least one 2.
What is the probability of seeing at least one 2 in any one roll of the pair of dice?
The probability of seeing at least one 2 = 1 - probability of seeing no 2s at all
So i calculate the probability of seeing no 2s at all.
x = 0 p = 1/6 q = 5/6 n = 10
Using the formula, p(x=0) = 10C0 * (1/6)^0 * (5/6)^10 Which gives me, p(x=0) = 0.1615
Probability of seeing at least one 2 = 1 - 0.1615 = 0.83
And i checked the ans, it stated the ans was 0.306.
So where did i go wrong? If someone can help me please, i would be much grateful