what is the probability of not throwing 7 or doubles for 6 consecutives throws with a pair of dice?

help please?!?!!

bruh

To find the probability of not throwing a 7 or doubles for 6 consecutive throws with a pair of dice, we need to break down the problem into smaller probabilities.

First, let's calculate the probability of not throwing a 7 on a single roll of the dice. The sum of throwing two dice ranges from 2 to 12, and there are 6 possible ways to get a 7. So, the probability of rolling a 7 is 6/36 or 1/6. Therefore, the probability of not rolling a 7 on a single roll is 1 - 1/6, which is 5/6.

Next, let's calculate the probability of not throwing doubles on a single roll. For any given roll, the probability of rolling doubles is 6/36 or 1/6 since there are 6 ways to roll doubles and 36 possible outcomes. Hence, the probability of not rolling doubles on a single roll is 1 - 1/6, which is also 5/6.

Now, we can calculate the probability of not throwing a 7 or doubles on a single roll. Since these events (not rolling a 7 and not rolling doubles) are mutually exclusive, we can multiply their individual probabilities: (5/6) * (5/6) = 25/36.

To find the probability for 6 consecutive throws, we multiply the probabilities together 6 times, as the throws are independent events. Therefore, the probability of not throwing a 7 or doubles for 6 consecutive throws is (25/36)^6, which is approximately 0.0137 or 1.37%.

So, the probability of not throwing a 7 or doubles for 6 consecutive throws with a pair of dice is about 1.37%.