Question

you pick a card at random, put it back, and then pick another card at random.
1 2 3 4
5 6
what is the probability of picking a divisor of 80 and then picking a 1?
simplify your answer and write it as a fraction or whole number.
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Explanation:

Step1: Count total cards

There are 6 cards (1,2,3,4,5,6).

Step2: Find divisors of 80

Divisors of 80 among the cards: 1,2,4,5 (since 80÷1=80, 80÷2=40, 80÷4=20, 80÷5=16; 3 and 6 don't divide 80 evenly). So 4 divisors.

Step3: Probability of divisor first

Probability = number of divisors / total cards = $\frac{4}{6} = \frac{2}{3}$.

Step4: Probability of picking 1 next

There is 1 card labeled 1. Probability = $\frac{1}{6}$.

Step5: Multiply probabilities (independent events)

Total probability = $\frac{2}{3} \times \frac{1}{6} = \frac{2}{18} = \frac{1}{9}$.

Answer:

$\frac{1}{9}$