Conrad Beltran

2022-09-18

Probability of Combinations which have different probabilties.

I cannot figure this out. The problem goes A boy has a bag filled with balls:

1 red

2 green

4 Blue

7 White

A child plays a game in which he withdraws one ball, writes down its color, and then replaces the ball. In order to win the game he must write down at least one of each color of ball. What is the probability that this will happen on, or before the N-th trial???

I cannot figure this out. The problem goes A boy has a bag filled with balls:

1 red

2 green

4 Blue

7 White

A child plays a game in which he withdraws one ball, writes down its color, and then replaces the ball. In order to win the game he must write down at least one of each color of ball. What is the probability that this will happen on, or before the N-th trial???

Haylee Krause

Beginner2022-09-19Added 10 answers

Step 1

This is 1 minus the probability that after N trials, one or more of the colours is missing.

So let us find the probability that some colour is missing. We will use the Method of Inclusion/Exclusion.

The probability red is missing is ${\left(\frac{14}{15}\right)}^{N}$.

We can write down similar expressions for the probability green is missing, blue is missing, white is missing.

Step 2

If we add up these 4 probabilities, we will have double-counted the situations in which 2 of the colours are missing. So we must subtract the probability that red and green is missing, together with 5 other similar expressions. The probability that red and green is missing is, for example, ${\left(\frac{12}{15}\right)}^{N}$.

But we have subtracted too much, for we have subtracted once too many times the situations in which 3 colours are missing. So we must add back 4 terms that represent these probabilities.

This is 1 minus the probability that after N trials, one or more of the colours is missing.

So let us find the probability that some colour is missing. We will use the Method of Inclusion/Exclusion.

The probability red is missing is ${\left(\frac{14}{15}\right)}^{N}$.

We can write down similar expressions for the probability green is missing, blue is missing, white is missing.

Step 2

If we add up these 4 probabilities, we will have double-counted the situations in which 2 of the colours are missing. So we must subtract the probability that red and green is missing, together with 5 other similar expressions. The probability that red and green is missing is, for example, ${\left(\frac{12}{15}\right)}^{N}$.

But we have subtracted too much, for we have subtracted once too many times the situations in which 3 colours are missing. So we must add back 4 terms that represent these probabilities.

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