Probability

(41) You have a deck of 52 cards. What is the probability of drawing a card that is neither a heart nor a spade?

• (a) \(\frac{1}{4}\)
• (b) \(\frac{26}{52}\)
• (c) \(\frac{25}{52}\)
• (d) \(\frac{27}{52}\)

Answer

Correct Answer: (d) \(\frac{27}{52}\)
P(\text{neither heart nor spade}) = 1 - P(\text{heart}) - P(\text{spade})
P(\text{heart}) = \frac{13}{52}
P(\text{spade}) = \frac{13}{52}
P(\text{neither heart nor spade}) = 1 - \frac{13}{52} - \frac{13}{52} = \frac{27}{52}.

(42) A bag contains 5 red marbles, 4 green marbles, and 3 blue marbles. If you draw three marbles without replacement, what is the probability that all three marbles are different colors?

• (a) \(\frac{10}{36}\)
• (b) \(\frac{15}{36}\)
• (c) \(\frac{60}{220}\)
• (d) \(\frac{12}{36}\)

Answer

Correct Answer: (c) \(\frac{60}{220}\)
P(\text{all different colors}) = \frac{\text{Number of ways to choose 3 different colors}}{\text{Total number of ways to choose 3 marbles}}
P(\text{all different colors}) = \frac{3C1 \times 4C1 \times 5C1}{12C3} = \frac{60}{220} = \frac{15}{55} = \frac{3}{11}.

(43) You have a jar with 10 candies: 6 are chocolate, and 4 are caramel. If you take two candies at random with replacement, what is the probability that both candies are caramel?

• (a) \(\frac{4}{15}\)
• (b) \(\frac{16}{100}\)
• (c) \(\frac{8}{25}\)
• (d) \(\frac{16}{225}\)

Answer

Correct Answer: (d) \(\frac{16}{225}\)
P(\text{both candies are caramel}) = P(\text{caramel}) \times P(\text{caramel})
P(\text{caramel}) = \frac{4}{10} = \frac{2}{5}
P(\text{both candies are caramel}) = \left(\frac{2}{5}\right) \times \left(\frac{2}{5}\right) = \frac{4}{25} = \frac{16}{100} = \frac{16}{225}.

(44) You have a bag with 8 red balls and 6 blue balls. If you draw one ball at random and then draw a second ball without replacement, what is the probability that both balls are blue?

• (a) \(\frac{6}{14}\)
• (b) \(\frac{3}{14}\)
• (c) \(\frac{3}{13}\)
• (d) \(\frac{3}{91}\)

Answer

Correct Answer: (d) \(\frac{3}{91}\)
P(\text{both balls are blue}) = P(\text{blue}) \times P(\text{blue after blue})
P(\text{blue}) = \frac{6}{14}
P(\text{blue after blue}) = \frac{5}{13}
P(\text{both balls are blue}) = \left(\frac{6}{14}\right) \times \left(\frac{5}{13}\right) = \frac{30}{182} = \frac{3}{91}.

(45) A box contains 4 red balls, 3 green balls, and 5 yellow balls. If you draw two balls without replacement, what is the probability that both balls are not red?

• (a) \(\frac{28}{55}\)
• (b) \(\frac{44}{55}\)
• (c) \(\frac{16}{33}\)
• (d) \(\frac{8}{15}\)

Answer

Correct Answer: (b) \(\frac{44}{55}\)
P(\text{both not red}) = P(\text{not red}) \times P(\text{not red after not red})
P(\text{not red}) = \frac{8}{12}
P(\text{not red after not red}) = \frac{7}{11}
P(\text{both not red}) = \left(\frac{8}{12}\right) \times \left(\frac{7}{11}\right) = \frac{56}{132} = \frac{44}{55}.