Question

elizabeth brought a box of donuts to share. there are two - dozen (24) donuts in the box, all identical in size, shape, and color. five are jelly - filled, 4 are lemon - filled, and 15 are custard - filled. you randomly select one donut, eat it, and select another donut. find the probability of selecting two custard - filled donuts in a row. (type an integer or a simplified fraction.)

Explanation:

Step1: Probability of first custard donut

There are 15 custard - filled donuts out of 24 total donuts. So the probability of selecting a custard - filled donut first, \(P_1=\frac{15}{24}\). We can simplify \(\frac{15}{24}=\frac{5}{8}\) (by dividing numerator and denominator by 3), but we can also work with the original fraction for now.

Step2: Probability of second custard donut (without replacement)

After selecting one custard - filled donut, there are now \(15 - 1=14\) custard - filled donuts left, and the total number of donuts left is \(24 - 1 = 23\). So the probability of selecting a custard - filled donut second, given that the first one was custard - filled, \(P_2=\frac{14}{23}\).

Step3: Probability of both events

Since we want the probability of selecting a custard - filled donut first AND then a custard - filled donut second, we use the multiplication rule for dependent events: \(P = P_1\times P_2\).

Substitute \(P_1=\frac{15}{24}\) and \(P_2=\frac{14}{23}\) into the formula:

\(P=\frac{15}{24}\times\frac{14}{23}=\frac{15\times14}{24\times23}=\frac{210}{552}\)

Simplify the fraction by dividing numerator and denominator by 6: \(\frac{210\div6}{552\div6}=\frac{35}{92}\)

Answer:

\(\frac{35}{92}\)