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Answer :
Suppose that we randomly choose pens from a box that contains yellow and green pens until a yellow pen is obtained. Assume that each selected pen is replaced before the next one is drawn. In order to find the probability of picking up a pen at least three times, we need to use the probability formula.
For this problem, the probability of choosing a yellow pen on the first draw is P(Y) = number of yellow pens / total number of pens = y / (y+g), where y is the number of yellow pens and g is the number of green pens. The probability of not choosing a yellow pen on the first draw is P(NY) = g / (y+g).After selecting a pen, if it is not yellow, we need to select a pen again. The probability of selecting a pen that is not yellow on the second draw is the same as the probability of not selecting a yellow pen on the first draw, that is[tex]P(NY) = g / (y+g).[/tex]
Therefore,
the probability of choosing a yellow pen on the second draw is P(Y and NY) = [tex]P(Y) × P(NY) = y × g / (y+g)²[/tex].The probability of not choosing a yellow pen on the first two draws is P(NYY) = P(NY) × P(NY) = g² / (y+g)².To calculate the probability of choosing a yellow pen on the third draw, we need to consider two cases: the pen selected on the first two draws is green, and the pen selected on the first two draws is not green.
Case 1: The pen selected on the first two draws is green. The probability of selecting a yellow pen on the third draw is P(Y and NY and G) =[tex]P(Y) × P(NY) × P(G) = y × g × (y+g) / (y+g)³.[/tex]
Case 2: The pen selected on the first two draws is not green. The probability of selecting a yellow pen on the third draw is P(Y and NY and NY) = P(Y) × P(NY) × P(NY) = y × g² / (y+g)³.Therefore, the probability of picking up a pen at least three times is P(at least 3) = P(NYY) + P(Y and NY and G) + P(Y and NY and NY) = g² / (y+g)² + y × g × (y+g) / (y+g)³ + y × g² / (y+g)³ = g² / (y+g)² + 2y × g / (y+g)³.
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