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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 Answers

6 votes

Answer:

2 successes = 80/243

0 successes = 32/243

at least 3 successes = 51/243

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

User Sajjad Sarkoobi
by
7.8k points
4 votes

Answer:

2 successes = 80/243

0 successes = 32/243

at least 3 successes = 51/243

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




User Zihadrizkyef
by
8.0k points

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