Question

A pair of fair dice is tossed 4 times.

Round your answer to 3 decimal places.

(a) Find the probability that 2 never occurs.
(b) Find the probability that 7 never occurs.

Answer 1

(a) .482
(b) .482

somebody check this cuz i'm not sure

Answer 2

how did you get that?

Answer 3

(5/6) ^ 4 because six sides 5 possiblities
available outcomes/ total outcomes

Answer 4

b was correct but a was not....

Answer 5

Are they independent events when you roll the dice each time? Yes! So you just multiply the probabilities together.

A)
You toss a pair of dice 4 times. So there are 4 events. You need to find P(not2)^4

Answer 6

Now, can we consider what P(not2) would be?

To roll a 2, you must roll a one on both dice, so P(2on two dice) = P(1) * P(1) = (1/6)*(1/6).

Now since we know P(2), the probability of not rolling a 2 is just the compliment, P(not2). And that is always equal to 1-P(2).

Answer 7

.972

Answer 8

So we have P(not2) = 1-(1/6)^2

And then raise that result to the 4th to get the likely hood of not rolling a 2 4 times in a row.

Answer 9

probability that no 2 occurs on on toss of pair of dice = 5/6 * 5/6 = 25/36
on 4 ththows this is 0.233

Answer 10

they are independent events

Answer 11

I think you're misinterpretting it, Jimbo. I think they care about the sum of a diceroll being 2.

Answer 12

probability of no 7 = 1 becuase there is no 7 on the dice!

Answer 13

Yeah, that's why I think you're misinterpretting it. Because obviously there is no 7 on the dice, but they ask about rolling a 7 as the next question. And why else would they word it as rolling a pair of dice 4 times instead of saying roll a die 8 times?

Answer 14

so is it just 1-(1/6)^2 which is .972 and then .972^4

Answer 15

Here's maybe a simpler way to understand it, Casemac. There are 36 possibilities for what comes up when you roll 2 dice. 6 possibilities for the first * 6 for the second = 36 possibilities.
There is only one way to roll a 2 though. So the probability of rolling a 2 is 1/36.

Then that means the probability of not rolling a 2 is 35/36 (Which happens to be 1-(1/36) which is the same as (1-(1/6)^2))

Answer 16

So if you roll 2 dice once, hoping not to get a 2, you have a 35/36 chance of success. If you do that 4 times, you have a chance of success of (35/36)*(35/36)*(35/36)*(35/36)

Answer 17

So yeah:
"so is it just 1-(1/6)^2 which is .972 and then .972^4"
That.