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.

Diego Barr
2022-10-20
Answered

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Claire Love

Answered 2022-10-21
Author has **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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What I've concluded : the first point A has range from [0;R] and the second point B must have range from $[0;R-OA]$ .But how should I find the probability?

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What I've concluded : the first point A has range from [0;R] and the second point B must have range from $[0;R-OA]$ .But how should I find the probability?

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Find the probability mass function of X"

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Find the probability mass function of X"

My thought is that x cannot be less than 2, since he would have to knock on two doors to make two sales.

I'm thinking the function would be ${\textstyle (}\genfrac{}{}{0ex}{}{x}{2}{\textstyle )}({u}^{2})(1-u{)}^{x-2}.$

But when I go to find E(x), that doesn't lend itself well to the geometric form I've learned to love.

Am I on the right track at least?

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