Question

you pick a card at random, put it back, and then pick another card at random.
what is the probability of picking a factor of 35 and then picking a number greater than 1?
simplify your answer and write it as a fraction or whole number.

Explanation:

Step1: Find the number of factors of 35 among 1,2,3,4

The factors of 35 are 1,5,7,35. Among 1,2,3,4, the factor of 35 is 1. So the number of favorable outcomes for the first - pick is \(n_1 = 1\). The total number of cards is \(N=4\). The probability of picking a factor of 35 in the first draw, \(P_1=\frac{n_1}{N}=\frac{1}{4}\).

Step2: Find the number of numbers greater than 1 among 1,2,3,4

The numbers greater than 1 among 1,2,3,4 are 2,3,4. So the number of favorable outcomes for the second - pick is \(n_2 = 3\). The total number of cards is still \(N = 4\) (since we put the first card back). The probability of picking a number greater than 1 in the second draw, \(P_2=\frac{n_2}{N}=\frac{3}{4}\).

Step3: Use the formula for the probability of independent events

Since the two draws are independent events (putting the card back means the outcome of the first draw does not affect the outcome of the second draw), and for two independent events \(A\) and \(B\), \(P(A\cap B)=P(A)\times P(B)\). Here \(A\) is the event of picking a factor of 35 and \(B\) is the event of picking a number greater than 1. So \(P = P_1\times P_2\). Substitute \(P_1=\frac{1}{4}\) and \(P_2=\frac{3}{4}\) into the formula: \(P=\frac{1}{4}\times\frac{3}{4}\).

Answer:

\(\frac{3}{16}\)