SSAT Upper Level Math : How to find the probability of an outcome

Study concepts, example questions & explanations for SSAT Upper Level Math

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Example Questions

Example Question #11 : Probability

A card is drawn at random from a standard 52-card deck. The person drawing wins if the card he draws is a two, a three, or a black four. What are the odds against drawing a winning card?

Possible Answers:

Correct answer:

Explanation:

There are 10 winning cards: the 4 twos, the 4 threes, and the 2 black fours. There are subsequently 42 losing cards, so the odds against a winning card are the number of losing cards, 42, to the number of winning cards, 10. This can be reduced to 

or 21 to 5.

Example Question #1661 : Ssat Upper Level Quantitative (Math)

A card is drawn at random from a standard 52-card deck. The person drawing wins if the card she draws is a red four or a black nine. What are the odds against drawing a winning card?

Possible Answers:

Correct answer:

Explanation:

There are 4 winning cards: the 2 red fours and the 2 black nines. There are subsequently 48 losing cards, so the odds against a winning card are the number of losing cards, 48, to the number of winning cards, 4. This can be reduced to 

or 12 to 1.

Example Question #11 : How To Find The Probability Of An Outcome

John participates in a charity fundraiser in which he pays $1 to draw a card from a standard 52-card deck. If he draws the ace of spades, he wins $25; if he draws any other ace, he wins $5; if he draws any other card, he does not win. To the nearest cent, what is the expected value of the game to John?

Possible Answers:

Correct answer:

Explanation:

Since there is only one card out of 52 (ace of spades) that wins John the $25 prize, the probability of this happening is . The value of this outcome to John is $24 - the $25 prize minus the $1 he paid to play.

Since there are three cards out of 52 (ace of clubs, ace of diamonds, ace of hearts) that win John a $5 prize, the probability of this happening is . The value of this outcome to John is $4 - the $5 prize minus the $1 he paid to play.

Since there are 48 out of 52 cards that do not win John a prize, the probability of this happening is . The value of this outcome to John is , since he had to pay $1 to play.

To find an expected value of a game, multiply the probability of each outcome by its value, then add the products. The expected value of this game to John is therefore

 dollars, or

The expected value of the game to Johnny rounds to .

Example Question #32 : Data Analysis / Probablility

Sharon participates in a charity fundraiser in which she pays $2 to draw a card from a deck of cards; the deck is a standard 52-card deck plus the joker. If she draws the joker, she wins $25; if she draws the ace of spades, she wins $10; if she draws any other ace, she wins $7; if she draws any other card, she does not win. To the nearest cent, what is the expected value of the game to Sharon?

Possible Answers:

Correct answer:

Explanation:

Since there is only one card out of 53 (joker) that wins Sharon the $25 prize, the probability of this happening is . The value of this outcome to Sharon is $23 - the $25 prize minus the $2 she paid to play.

Since there is only one card out of 53 (ace of spades) that wins Sharon the $10 prize, the probability of this happening is . The value of this outcome to Sharon is $8 - the $10 prize minus the $2 she paid to play.

Since there are three cards out of 53 (ace of clubs, ace of diamonds, ace of hearts) that win Sharon a $7 prize, the probability of this happening is . The value of this outcome to Sharon is $5 - the $7 prize minus the $2 she paid to play.

Since there are 48 out of 53 cards that do not win Sharon a prize, the probability of this happening is . The value of this outcome to Sharon is , since he had to pay $2 to play.

To find an expected value of a game, multiply the probability of each outcome by its value, then add the products. The expected value of this game to Sharon is therefore

  dollars, or

The expected value of the game to Sharon rounds to .

Example Question #11 : How To Find The Probability Of An Outcome

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?

Possible Answers:

50%

19.8%

27.5%

33%

6.6%

Correct answer:

27.5%

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 .

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

which is 27.5%.

Example Question #41 : How To Find The Probability Of An Outcome

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

Possible Answers:

Correct answer:

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 .

There are thirteen spades in the deck, so the probability of drawing a spade is .

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 

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

 

Example Question #45 : Probability

Set A: 

Set B: 

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

Possible Answers:

Correct answer:

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 .

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 .

 

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:

 

Example Question #46 : Probability

Set A: 

Set B: 

One letter is picked from Set A and Set B. What is the probability of picking at least one consonant?

Possible Answers:

Correct answer:

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 .

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 .

The question asks for the probability of drawing at least one consonant, which can be interpreted as a union of events. To calculate the probability of a union, sum the probability of each event and subtract the intersection:

The interesection is:

So, we can find the probability of drawing at least one consonant:

 

Example Question #42 : Probability

Set A: 

Set B: 

One letter is drawn from Set A, and one from Set B. What is the probability of drawing a matching pair of letters?

Possible Answers:

Correct answer:

Explanation:

Set A: 

Set B: 

Between Set A and Set B, there are two potential matching pairs of letters: AA and XX. The amount of possible combinations is the number of values in Set A, multiplied by the number of values in Set B, .

Therefore, the probability of drawing a matching set is:

Example Question #51 : How To Find The Probability Of An Outcome

In a particular high school, 200 students are freshmen, 150 students are sophomores, 250 students are juniors, and 100 students are seniors. Twenty percent of freshmen are in honors classes, ten percent of sophomores are in honors classes, twelve percent of juniors are in honors classes, and thirty percent of seniors are in honors classes.

If a student is chosen at random, what is the probability that that student will be a student who attends honors classes?

Possible Answers:

Correct answer:

Explanation:

First calculate the number of students:

 

The probability of drawing an honors student will then be the total number of honors students divided by the total number of students attending the school:

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