Probability

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SSAT Upper Level Quantitative › Probability

Questions 1 - 10

In a jar, there are 3 blue marbles, 5 red marbles, 8 green marbles. If Bob reaches his hand in a jar, and grabs one marble, what is the likelihood he will pick up a blue marble?

$\frac{3}{16}$

$\frac{3}{13}$

$\frac{3}{8}$

$\frac{3}{5}$

$\frac{5}{11}$

Explanation

First, calculate the total number of marbles in the jar, which is . Because 3 of the marbles are blue, the chances of picking a blue marble $\frac{3}{16}$ .

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

In a jar, there are 3 blue marbles, 5 red marbles, 8 green marbles. If Bob reaches his hand in a jar, and grabs one marble, what is the likelihood he will pick up a blue marble?

$\frac{3}{16}$

$\frac{3}{13}$

$\frac{3}{8}$

$\frac{3}{5}$

$\frac{5}{11}$

Explanation

First, calculate the total number of marbles in the jar, which is . Because 3 of the marbles are blue, the chances of picking a blue marble $\frac{3}{16}$ .

In a jar, there are 3 blue marbles, 5 red marbles, 8 green marbles. If Bob reaches his hand in a jar, and grabs one marble, what is the likelihood he will pick up a blue marble?

$\frac{3}{16}$

$\frac{3}{13}$

$\frac{3}{8}$

$\frac{3}{5}$

$\frac{5}{11}$

Explanation

First, calculate the total number of marbles in the jar, which is . Because 3 of the marbles are blue, the chances of picking a blue marble $\frac{3}{16}$ .

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

Set A:

Set B:

One letter is picked from Set A and Set B. What is the probability of picking two consonants?

$\frac{5}{14}$

$\frac{4}{7}$

$\frac{5}{7}$

$\frac{2}{14}$

$\frac{3}{14}$

Explanation

Set A:

Set B:

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

The question asks for the probability of drawing two consonants, meaning the probability of drawing a constant from Set A and Set B, so probability of the intersection of the two events is the product of the two probabilities:

$\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

Set A:

Set B:

One letter is picked from Set A and Set B. What is the probability of picking two consonants?

$\frac{5}{14}$

$\frac{4}{7}$

$\frac{5}{7}$

$\frac{2}{14}$

$\frac{3}{14}$

Explanation

Set A:

Set B:

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

$\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

Set A:

Set B:

One letter is picked from Set A and Set B. What is the probability of picking two consonants?

$\frac{5}{14}$

$\frac{4}{7}$

$\frac{5}{7}$

$\frac{2}{14}$

$\frac{3}{14}$

Explanation

Set A:

Set B:

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

$\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

Jeff collects basketball cards of players on his three favorite teams. He decides to put 5 cards from each team in a paper bag and then to draw out 3 cards at random. What are the odds of him getting one player from each team?

27.5%

6.6%

19.8%

33%

50%

Explanation

For this problem, we will multiply together the odds of each draw (assuming he draws one card at a time...the odds won't change if he draws three at once, but it's easier to visualize this way) resulting in a card that works for Jeff's goal of having one player from each team. The first draw cannot fail, as he needs one player from each team and the first card he draws must be from one of the teams. After this draw, he has 14 cards remaining, and 10 of these are players on the two teams that can still offer a player.

So the odds of a successful second draw are $\frac{5}{7}$ .

The last draw is the trickiest, as there would now be 13 cards remaining, with only 5 being players from the team that he still needs represented. When we multiply all of these odds together, we get

$\frac{1}{1}\times \frac{5}{7}\times\frac{5}{13}$

which is 27.5%.

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