Question

you pick a card at random. without putting the first card back, you pick a second card at random.
2 3 4 5
6 7 8
what is the probability of picking an odd number and then picking an odd number?
write your answer as a fraction or whole number.

Explanation:

Step1: Count total and odd cards

Total cards: \( 4 + 3 = 7 \). Odd cards (3, 5, 7): \( 3 \). Probability first odd: \( \frac{3}{7} \).

Step2: Probability second odd (no replacement)

After 1 odd card, total cards: \( 6 \), odd cards: \( 2 \). Probability second odd: \( \frac{2}{6}=\frac{1}{3} \).

Step3: Multiply probabilities

Probability both odd: \( \frac{3}{7} \times \frac{1}{3} = \frac{1}{7} \). Wait, correction: Wait, total cards: 7? Wait, first row 4, second row 3: total 7. Odd cards: 3 (3,5,7). After picking one odd, remaining total: 6, remaining odd: 2. So \( \frac{3}{7} \times \frac{2}{6} = \frac{6}{42} = \frac{1}{7} \)? Wait no: \( 3\times2=6 \), \( 7\times6=42 \), simplify \( \frac{6}{42}=\frac{1}{7} \)? Wait, no: 3/7 2/6 = (32)/(76) = 6/42 = 1/7. Wait, but let's recheck total cards: 2,3,4,5,6,7,8: that's 7 cards. Odd numbers: 3,5,7: 3. So first pick: 3/7. Then, remaining cards: 6, odd remaining: 2. So second pick: 2/6 = 1/3. Multiply: 3/7 1/3 = 1/7. Yes.

Answer:

\(\frac{1}{7}\)