# Thread: probability of dice throw

1. ## probability of dice throw

Question: A die has 6 sides numbered 1-6. If 3 dices are thrown, what is the probability that at least one of the dices shows the number six?

For this question, do I have to take into consideration that the other two dices are not sixes? This is what I get:

1/6 for first die to be a 6
(5/6)*(1/6) for second die to be a 6 if the first is not
(25/36)*(1/6) for the third die to be a 6 if the first two are not

adding them up I get 91/216

Is my analysis correct?

2. Hello, chemekydie!

A die has 6 sides numbered 1-6.
If 3 dices are thrown, what is the probability that
at least one of the dices shows the number six?

For this question, do I have to take into consideration that the other two dices are not sixes?

This is what I get:

$\frac{1}{6}$ for first die to be a 6

$\frac{5}{6}\cdot\frac{1}{6}$ for second die to be a 6 if the first is not.

$\frac{25}{36}\cdot\frac{1}{6}$ for the third die to be a 6 if the first two are not.

Adding them up I get: $\frac{91}{216}$

Is my analysis correct?

Yes! . . . Good work!

There is a back-door approach to this problem.

The opposite of "at least one 6" is "no 6's."

The probability of no 6's is: . $\left(\frac{5}{6}\right)^3 \:=\:\frac{125}{216}$

Therefore: . $P(\text{at least one 6}) \;=\;1 - \frac{125}{216} \;=\;\frac{91}{216}$