You will add the two numbers together and add the t as well to make 33t
The answer to your question is 215
Answer:
23.44%
Step-by-step explanation:
The probability of getting a 4 on the first 2 throws and different numbers on the last 5 throws = 1/6 * 1/6 * (5/6)^5
= 0.01116
There are 7C2 ways of the 2 4's being in different positions
= 7*6 / 2 = 21 ways.
So the required probability = 0.01116 * 21
= 0.2344 or 23.44%.
