Two baskets C and D each contain a mixture of oranges and lemons. Basket C contains 26 oranges and 13 lemons. Basket D contains 18 oranges and 15 lemons. A child selected basket at random and picked at random a fruit from it. Determine the probability that the fruit picked was an orange.
success could be :
basketC, orange
basketD, orange
prob = (1/2)(26/39) + (1/2)(18/33)
= 20/33
To determine the probability of picking an orange, we need to calculate the total number of oranges and the total number of fruits in both baskets.
In basket C, the total number of fruits is 26 + 13 = 39.
In basket D, the total number of fruits is 18 + 15 = 33.
Therefore, the total number of fruits in both baskets is 39 + 33 = 72.
The probability of selecting an orange is the number of oranges divided by the total number of fruits. So, the probability is (26 + 18) / 72 = 44 / 72.
To simplify this probability, we can find the greatest common divisor (GCD) of 44 and 72, which is 4. Dividing both numerator and denominator by 4 gives us 11 / 18.
Therefore, the probability of picking an orange is 11 / 18 or approximately 0.6111, which is about 61.11%.
To determine the probability that the fruit picked was an orange, we need to know the total number of fruits in both baskets and the number of oranges.
In basket C, there are 26 oranges and 13 lemons, so the total number of fruits in basket C is 26 + 13 = 39.
In basket D, there are 18 oranges and 15 lemons, so the total number of fruits in basket D is 18 + 15 = 33.
The total number of fruits in both baskets is 39 + 33 = 72.
Therefore, the probability of picking an orange is the number of oranges divided by the total number of fruits:
P(orange) = (26 + 18) / 72
= 44 / 72
Simplifying this fraction gives the final answer:
P(orange) = 11 / 18
So, the probability that the fruit picked was an orange is 11/18.