Question

"What is the probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice? Assume you’re throwing a single die each time."

Answer 1

Answer:

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

Step-by-step explanation:

For each throw, there are only two possible outcomes. Either it is a '3', or it is not. This means that we solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

There are 6 possible outcomes for the dice. This means that the probability that it is a '3' is $\frac{1}{6} = 0.167$

There are 10 throws, so $n = 10$.

Probability of throwing AT LEAST two ‘3’s on the dice?

Either you throw less than two, or you throw at least two. The sum of the probabilities of these events is 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X \geq 2) = 1 - P(X < 2)$.

In which

$P(X < 2) = P(X = 0) + P(X = 1)$.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.167)^{0}.(0.833)^{10} = 0.1609$

$P(X = 1) = C_{10,1}.(0.167)^{1}.(0.833)^{9} = 0.3225$

So

$P(X < 2) = P(X = 0) + P(X = 1) = 0.1609 + 0.3225 = 0.4834$.

Finally:

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4839 = 0.5161$.

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.