Gardiolo0j

2022-09-03

How can i solve this question with Geometric distribution or random variables?

I tried have tried using a 'Geometric distribution', but it hasn't worked.

John and Ron play basketball 10 times. The probability that John wins in single round $=0.4$. The probability that Ron wins in single round $=0.3$. The probability that there is equality between them in a single round $=0.3$. The "Winner" is defined to be the first to win a single round.

The rotations are different and independent.

What is the probability that John is the Winner?

I tried have tried using a 'Geometric distribution', but it hasn't worked.

John and Ron play basketball 10 times. The probability that John wins in single round $=0.4$. The probability that Ron wins in single round $=0.3$. The probability that there is equality between them in a single round $=0.3$. The "Winner" is defined to be the first to win a single round.

The rotations are different and independent.

What is the probability that John is the Winner?

Rihanna Blanchard

Beginner2022-09-04Added 13 answers

Step 1

Given that there are exactly 10 rounds, the probability of having no winner by the end of the 10 rounds is 0.310. Then we know that the probability of having a winner is $1-{0.3}^{10}$.

Step 2

Such probability should be apportioned between John and Ron in the ratio 0.4/0.3. So the probability that John wins is $\frac{4}{7}(1-{0.3}^{10})$

Given that there are exactly 10 rounds, the probability of having no winner by the end of the 10 rounds is 0.310. Then we know that the probability of having a winner is $1-{0.3}^{10}$.

Step 2

Such probability should be apportioned between John and Ron in the ratio 0.4/0.3. So the probability that John wins is $\frac{4}{7}(1-{0.3}^{10})$

ohgodamnitw0

Beginner2022-09-05Added 2 answers

Explanation:

Since they only play 10 times, the probability that John wins is equal to the sum of probabilities that John wins in the i-th round:

$P=\sum _{i=1}^{10}({0.3}^{i-1}\times 0.4)=0.4\times \frac{1-{0.3}^{10}}{0.7}$ where the probability that John wins in the i-th round is ${0.3}^{i-1}\times 0.4$ because the first $i-1$ rounds were all draws.

Since they only play 10 times, the probability that John wins is equal to the sum of probabilities that John wins in the i-th round:

$P=\sum _{i=1}^{10}({0.3}^{i-1}\times 0.4)=0.4\times \frac{1-{0.3}^{10}}{0.7}$ where the probability that John wins in the i-th round is ${0.3}^{i-1}\times 0.4$ because the first $i-1$ rounds were all draws.

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