Red is The New Green

Solution

Let \(R\) denote drawing a red ball and \(G\) denote drawing a green ball. Initially, the bag contains \(14 + 6 = 20\) balls, so \[ P(R) = \frac{14}{20} = 0.7 \qquad \text{and} \qquad P(G) = \frac{6}{20} = 0.3. \] We want the probability of drawing exactly \(3\) red balls in \(5\) draws. For both parts, the number of ways to choose which \(3\) of the \(5\) draws are red is \[ \binom{5}{3} = 10. \] Case 1: With replacement Because the ball is replaced after each draw, each draw is independent, and the probability of red remains \(0.7\) on every draw. Thus, the number of red balls in \(5\) draws follows a binomial distribution with parameters \(n=5\) and \(p=0.7\). Therefore, \[ P(\text{exactly }3\text{ red}) = \binom{5}{3}(0.7)^3(0.3)^2. \] Substituting in values, \[ = 10 \cdot (0.7)^3 \cdot (0.3)^2 \] \[ = 10 \cdot 0.343 \cdot 0.09 \] \[ = 10 \cdot 0.03087 \] \[ = 0.3087. \] So, with replacement, the probability is \( 0.3087 \approx 30.87\%. \)

Case 2: Without replacement Now the draws are no longer independent, since removing a ball changes the composition of the bag. We use the hypergeometric model. Out of the \(20\) balls, we want exactly \(3\) reds from the \(14\) red balls and exactly \(2\) greens from the \(6\) green balls. The number of favourable samples is \[ \binom{14}{3}\binom{6}{2}. \] The total number of ways to choose any \(5\) balls from \(20\) is \[ \binom{20}{5}. \] Therefore, \[ P(\text{exactly }3\text{ red}) = \frac{\binom{14}{3}\binom{6}{2}}{\binom{20}{5}}. \] \[ P(\text{exactly }3\text{ red}) = \frac{364 \cdot 15}{15504} \] \[ = \frac{5460}{15504} \] \[ \approx 0.3522. \] So, without replacement, the probability is \( 0.3522 \approx 35.22\%. \) Thus, drawing without replacement gives a slightly higher probability of obtaining exactly \(3\) red balls in this case.