Question

a box has 4 red and 6 blue balls. one ball is drawn at random and replaced, then a second ball is drawn. what is p(both red)? answer using a fraction in simplest form.

Explanation:

Step1: Find total number of balls

There are 4 red and 6 blue balls, so total balls = \(4 + 6 = 10\).

Step2: Probability of first red ball

Probability of drawing a red ball first is \(\frac{\text{number of red balls}}{\text{total number of balls}}=\frac{4}{10}\).

Step3: Probability of second red ball (with replacement)

Since we replace the ball, the probability of drawing a red ball second is also \(\frac{4}{10}\).

Step4: Probability of both red (independent events)

For independent events, \(P(A \text{ and } B)=P(A)\times P(B)\). So \(P(\text{both red})=\frac{4}{10}\times\frac{4}{10}=\frac{16}{100}\). Simplify this fraction by dividing numerator and denominator by 4: \(\frac{16\div4}{100\div4}=\frac{4}{25}\).

Answer:

\(\frac{4}{25}\)