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a basket has 9 fruits: 3 apples, 6 oranges. 1 fruit is randomly drawn a…

Question

a basket has 9 fruits: 3 apples, 6 oranges. 1 fruit is randomly drawn and eaten, then another 1 is drawn. what is the probability of drawing an apple then an orange?
event a: apple on 1st draw
event b: orange on 2nd draw
p(a, then b) =?
compound probability
dependent events
p(a, then b) =
p(a) - p(b after a)
give your answer as a fraction in simplest form.

Explanation:

Step1: Calculate P(A)

Probability of apple first: $\frac{3}{9} = \frac{1}{3}$

Step2: Calculate P(B after A)

After 1 apple drawn, 8 fruits left (2 apples, 6 oranges). Probability of orange: $\frac{6}{8} = \frac{3}{4}$

Step3: Multiply probabilities

$P(A \text{ then } B) = \frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$

Answer:

$\frac{1}{4}$