Someone rolls a six sided fair die 5 times. they consider it a success if they have a two or three on a die. whats the probability of 2 successes? no successes? and at least 3 successes?

2 successes =  80/243

0 successes = 32/243

at least 3 successes = 51/243

Step-by-step explanation:

Lets suppose we had the following sequence of 5 throws:-

2 or 3,  2 or 3 ,  then NOT 2 or 3 on last 3 throws

Probability of this is

1/3 * 1/3 * 2/3 * 2/3 * 2/3 = 8/243

There are  (5*4) / 2 = 10 ways of having this result  ( that is the number of combinations of 2 from  5   =  5C2 ).

So Prob(2 successes) = (8/243) * 10  = 80/243

Prob ( No successes) =  (2/3)^5 = 32/243

Probability of 3 successes = 1/3 * 1/3 * 1/3 * 2/3 * 2/3 *   5C3  

4/243 * 10 = 40 / 243

Probability of 4 successes =  (1/3)^4 * 2/3 * 5C4

= 10/243

Probability of 5 success = (1/3)^5 = 1 / 243

Prob(at least 3 successes) = 51/243

im guessing (23,14). hope this : )