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35 birds are in a cage. 14 of them are robins. a randomly selected bird…

Question

35 birds are in a cage. 14 of them are robins. a randomly selected bird is taken from the cage. without replacing the first one, a second bird is chosen. what is the probability that both of the birds chosen are robins? round your answer to the nearest hundredth

Explanation:

Step1: Calculate first - pick probability

The probability of choosing a robin on the first pick is the number of robins divided by the total number of birds. There are 14 robins and 35 total birds, so the probability $P_1=\frac{14}{35}=\frac{2}{5}$.

Step2: Calculate second - pick probability

After one robin is chosen, there are 13 robins left and 34 birds left. So the probability of choosing a robin on the second pick given that a robin was chosen on the first pick is $P_2 = \frac{13}{34}$.

Step3: Calculate the joint probability

Since these are dependent events, the probability that both birds are robins is the product of the probabilities of each event. So $P = P_1\times P_2=\frac{2}{5}\times\frac{13}{34}=\frac{26}{170}\approx 0.15$.

Answer:

0.15