Calculating the Number of Ways to Select 6 Red Balls from an Urn with Replacement

Imagine you have an urn containing 5 red balls and 3 blue balls. You wish to select 9 balls from the urn with replacement, with the goal of selecting exactly 6 red balls. This problem requires an understanding of combinations with replacement and multinomial coefficients, which are crucial in solving problems related to probability and combinatorics.

Understanding the Problem

The problem at hand involves selecting 6 red balls out of 9 draws, with the urn containing only 5 red balls and 3 blue balls. Since the balls are drawn with replacement, the probability of drawing a red ball remains constant at each draw. We need to determine how many possible ways there are to draw 6 red balls and 3 blue balls in 9 draws.

Solution Through Combinations with Replacement

To solve this problem, we can use the concept of combinations with replacement. The key here is to recognize that we need to distribute the remaining 3 draws among the blue balls. Let's denote the number of blue balls selected as ( b ). Since we are drawing a total of 9 balls, we can express this as:

( r b 9 )

Where ( r 6 ) is the number of red balls.

Substituting the value of ( r ):

( 6 b 9 ) implies ( b 3 )

We need to select 3 blue balls. The draws are independent due to the replacement. The number of ways to choose 6 red balls and 3 blue balls is given by the multinomial coefficient:

( frac{9!}{6! cdot 3!} )

Let's calculate this step by step:

First, we calculate the factorials:

( 9! 362880 )

( 6! 720 )

( 3! 6 )

Now, substituting these values into the formula:

( frac{362880}{720 cdot 6} frac{362880}{4320} 84 )

Therefore, the number of ways to select 6 red balls when drawing 9 balls with replacement is 84.

Alternative Method: Permutations with Repetition

Another way to approach this problem is to consider it as a permutation problem involving 9 objects with 6 of one type (red balls) and 3 of another (blue balls). The formula for such a problem is:

( frac{9!}{6! cdot 3!} )

Calculating this:

( 9! 362880 )

( 6! 720 )

( 3! 6 )

Substituting these values:

( frac{362880}{720 cdot 6} frac{362880}{4320} 84 )

Hence, there are 84 ways to draw 6 red balls and 3 blue balls in 9 draws with replacement.

Probability Considerations

While the OP is not interested in probabilities but in the number of ways to draw balls of a specific color after a specific number of draws, let's explore the probability aspect for educational purposes:

The probability of drawing a red ball is ( p frac{5}{8} ), and the probability of drawing a blue ball is ( q 1 - p frac{3}{8} ).

Considering ( x 6 ) (success - drawing a red ball) and ( n 9 ) (total number of trials), the formula for binomial distribution is:

( P(X r) binom{n}{r} p^x q^{n-r} )

Plugging in the values:

( P(X 6) binom{9}{6} left( frac{5}{8} right)^6 left( frac{3}{8} right)^3 )

Calculating this:

( binom{9}{6} 84 )

( left( frac{5}{8} right)^6 approx 0.0238 )

( left( frac{3}{8} right)^3 approx 0.0527 )

Therefore:

( 84 cdot 0.0238 cdot 0.0527 approx 0.13516 )

This means that the probability of drawing 6 red balls in 9 trials is approximately 0.135, or 135 ways when considering the number of trials.

While the probability aspect is interesting, the key takeaway is the number of ways to draw 6 red balls, which is 84, without considering the probabilities of success and failure being equal.