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a bag contains 15 candies: 6 red, 9 blue. if 1 is randomly drawn and ea…

Question

a bag contains 15 candies: 6 red, 9 blue. if 1 is randomly drawn and eaten, then another 1 is drawn, what is the probability of drawing a blue candy then red candy? event a: blue candy on 1st draw event b: red candy on 2nd draw p(a, then b) = ?/? give your answer as a fraction in simplest form compound probability dependent events p(a, then b) = p(a) · p(b after a)

Explanation:

Step1: Calculate P(A)

P(A) = number of blue candies / total candies = $\frac{9}{15}$ = $\frac{3}{5}$

Step2: Calculate P(B after A)

After drawing blue, total candies = 14, red candies = 6. P(B after A) = $\frac{6}{14}$ = $\frac{3}{7}$

Step3: Compute P(A then B)

Multiply P(A) and P(B after A): $\frac{3}{5} \times \frac{3}{7}$ = $\frac{9}{35}$

Answer:

$\frac{9}{35}$