# A dice is rolled, what is the probability of getting 1 and 6?

## A. 1/6 B. 2/36 C. 1/36 D. 3/6

Mar 1, 2018

$\frac{2}{36}$

#### Explanation:

I assume that dice has been rolled twice. So, there could be 1 number on the top side each time.
I also assume that the numbers may or may not appear in given sequence.

probability of getting $1$ or $6$ first time $= \frac{2}{6}$

Now since we have got either of six or one we need the other one now.

Examples are better for explaining. So, consider that 1 appeared on top now we need 6.
So probability of getting $6 \text{ is } \frac{1}{6}$.
Multiply both probabilities and we get 2/6×1/6=2/36
(note we could have got 6 also first time but that doesn't hinder our calculation)

If we ask "What is the probability of rolling a 1 and then rolling a 6?", the answer is C $= \frac{1}{36}$

#### Explanation:

I'm going to run through what we're given and see if we can piece this together.

When we roll a single die, we're only going to get a single roll - we can't get both a 1 and a 6, which is what the question currently asks.

So what if we ask the question - What is the probability of rolling a 1 and then rolling a 6?

With this question, we have an answer that makes sense.

The probability of rolling a 1 is $\frac{1}{6}$. On the next roll, the probability of rolling a 6 is also $\frac{1}{6}$. Since these are independent events, we multiply to see the probability of rolling a 1 and then rolling a 6:

$\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

If we ask the question What is the probability of rolling a 1 or a 6, we look at the number of ways we can meet the conditions $\left(= 2\right)$ over the number of ways a die can come up $\left(= 6\right)$, and get:

$\frac{2}{6}$ - but there is no option for this and so probably isn't the question being asked.