Diego Barr

2022-10-20

Suppose we roll a fair dice repeatedly. Find the probability that we get our 4th prime number on the 25th roll. I think it should be 4 successes and 21 failures but I feel it incorrect.

Claire Love

Beginner2022-10-21Added 14 answers

Step 1

The prime numbers on a dice are 2,3,5. To calculate the probability that we get the 4th prime number on the 25th roll, we have to calculate the number of possibilities to role exactly four prime numbers in the first 24 rolls.

So the amount of ways to place exactly three prime numbers in 24 rolls.

This leads to ${3}^{3}\cdot {3}^{21}\cdot {\textstyle (}\genfrac{}{}{0ex}{}{24}{3}{\textstyle )}\cdot 3$

The first factor represents that we have to choose from 3 different prime numbers three times. Similar the second factor represents that we have to choose from three different non-prime numbers 21 times. The third factor (243) represents the different ways we can arrange the different prime numbers among the 24 rolls. (The might accour on the first, second and third turn, but could also accour on the first, 17th and 20th ... and so on. We have to take this in consideration).

Step 2

The last factor represents the 25th role, which has to be a prime number. For that we can again choose 2,3 or 5, so three possible outcomes.

Also the number of overall possible ways this "game" can end (throwing a dice 25 times in a row) is 625

This gives us the probability:

$\frac{{3}^{3}\cdot {3}^{21}\cdot {\textstyle (}\genfrac{}{}{0ex}{}{24}{3}{\textstyle )}\cdot 3}{{6}^{25}}$

For a full solution you should try to model the problem accordingly.

The prime numbers on a dice are 2,3,5. To calculate the probability that we get the 4th prime number on the 25th roll, we have to calculate the number of possibilities to role exactly four prime numbers in the first 24 rolls.

So the amount of ways to place exactly three prime numbers in 24 rolls.

This leads to ${3}^{3}\cdot {3}^{21}\cdot {\textstyle (}\genfrac{}{}{0ex}{}{24}{3}{\textstyle )}\cdot 3$

The first factor represents that we have to choose from 3 different prime numbers three times. Similar the second factor represents that we have to choose from three different non-prime numbers 21 times. The third factor (243) represents the different ways we can arrange the different prime numbers among the 24 rolls. (The might accour on the first, second and third turn, but could also accour on the first, 17th and 20th ... and so on. We have to take this in consideration).

Step 2

The last factor represents the 25th role, which has to be a prime number. For that we can again choose 2,3 or 5, so three possible outcomes.

Also the number of overall possible ways this "game" can end (throwing a dice 25 times in a row) is 625

This gives us the probability:

$\frac{{3}^{3}\cdot {3}^{21}\cdot {\textstyle (}\genfrac{}{}{0ex}{}{24}{3}{\textstyle )}\cdot 3}{{6}^{25}}$

For a full solution you should try to model the problem accordingly.

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