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a fruit is randomly drawn from a basket with 20 red apples, 6 green app…

Question

a fruit is randomly drawn from a basket with 20 red apples, 6 green apples, 14 green pears, and 10 yellow pears. event a: green event b: pear p(a or b) =? hint: p(a or b) = p(a) + p(b) - p(a and b) round your answer to the nearest hundredth

Explanation:

Step1: Calculate total number of fruits

$20 + 6+14 + 10=50$

Step2: Calculate P(A)

Number of green fruits is $6 + 14=20$. So $P(A)=\frac{20}{50}=0.4$

Step3: Calculate P(B)

Number of pears is $14 + 10=24$. So $P(B)=\frac{24}{50}=0.48$

Step4: Calculate P(A and B)

Number of green pears is $14$. So $P(A\ and\ B)=\frac{14}{50}=0.28$

Step5: Calculate P(A or B)

Using the formula $P(A\ or\ B)=P(A)+P(B)-P(A\ and\ B)$, we have $P(A\ or\ B)=0.4 + 0.48- 0.28=0.6$

Answer:

$0.60$