Probability

(46) A box contains 7 white balls, 4 black balls, and 3 red balls. If you draw one ball at random, what is the probability that it is either white or red?

• (a) \(\frac{7}{14}\)
• (b) \(\frac{4}{14}\)
• (c) \(\frac{10}{14}\)
• (d) \(\frac{11}{14}\)

Answer

Correct Answer: (c) \(\frac{10}{14}\)
P(\text{white or red}) = P(\text{white}) + P(\text{red})
P(\text{white}) = \frac{7}{14}
P(\text{red}) = \frac{3}{14}
P(\text{white or red}) = \frac{7}{14} + \frac{3}{14} = \frac{10}{14} = \frac{5}{7}.

(47) You have a standard deck of 52 cards. What is the probability of drawing a face card (jack, queen, or king) or a black card?

• (a) \(\frac{16}{52}\)
• (b) \(\frac{12}{52}\)
• (c) \(\frac{32}{52}\)
• (d) \(\frac{24}{52}\)

Answer

Correct Answer: (c) \(\frac{32}{52}\)
P(\text{face card or black card}) = P(\text{face card}) + P(\text{black card}) - P(\text{both})
P(\text{face card}) = \frac{12}{52}
P(\text{black card}) = \frac{26}{52}
P(\text{both}) = \frac{6}{52}
P(\text{face card or black card}) = \frac{12}{52} + \frac{26}{52} - \frac{6}{52} = \frac{32}{52} = \frac{8}{13}.

(48) In a bag, there are 9 blue marbles, 6 green marbles, and 5 red marbles. If you draw two marbles without replacement, what is the probability that both marbles are green?

• (a) \(\frac{6}{20}\)
• (b) \(\frac{15}{190}\)
• (c) \(\frac{30}{190}\)
• (d) \(\frac{9}{190}\)

Answer

Correct Answer: (b) \(\frac{15}{190}\)
P(\text{both green}) = P(\text{green}) \times P(\text{green after green})
P(\text{green}) = \frac{6}{20}
P(\text{green after green}) = \frac{5}{19}
P(\text{both green}) = \left(\frac{6}{20}\right) \times \left(\frac{5}{19}\right) = \frac{30}{380} = \frac{15}{190}.

(49) A box contains 12 chocolates, 8 candies, and 6 cookies. If you draw one treat at random, what is the probability that it is either chocolate or candy?

• (a) \(\frac{20}{26}\)
• (b) \(\frac{20}{26}\)
• (c) \(\frac{12}{26}\)
• (d) \(\frac{14}{26}\)

Answer

Correct Answer: (b) \(\frac{20}{26}\)
P(\text{chocolate or candy}) = P(\text{chocolate}) + P(\text{candy})
P(\text{chocolate}) = \frac{12}{26}
P(\text{candy}) = \frac{8}{26}
P(\text{chocolate or candy}) = \frac{12}{26} + \frac{8}{26} = \frac{20}{26} = \frac{10}{13}.

(50) You roll two fair six-sided dice. What is the probability that the sum of the numbers rolled is 7?

• (a) \(\frac{1}{6}\)
• (b) \(\frac{6}{36}\)
• (c) \(\frac{1}{36}\)
• (d) \(\frac{2}{6}\)

Answer

Correct Answer: (b) \(\frac{6}{36}\)
There are 6 ways to get a sum of 7 when rolling two dice: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).
P(\text{sum of 7}) = \frac{6}{36} = \frac{1}{6}.