How to find the probability of an outcome - SSAT Upper Level Quantitative

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Question

Ten balls - five green, three blue, two yellow - are placed in a hat. A ball is drawn, then replaced; then, a ball is drawn again. What is the probability that a blue ball will be drawn at least once?

Answer

This can more easily be solved by finding the probability that neither draw will result in a blue ball, then subtracting it from 1.

The probability that the first ball will not be blue will be seven (balls that are not blue) out of ten (balls total), or $\frac{7}{10}$ . Since the ball is replaced, this is also the probability that the second will not be blue. Therefore, the probability that neither will be blue is $\frac{7}{10} \times \frac{7}{10} = \frac{49}{100}$ .

The probability that at least one draw will result in a blue ball is therefore $1- \frac{49}{100} = \frac{100}{100} - \frac{49}{100} = \frac{51}{100}$ .

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Question

Ten balls - five green, three blue, two yellow - are placed in a hat. Two balls are drawn, one after the other, without replacement. What is the probability that at least one will be green?

Answer

This can more easily be solved by finding the probability that neither will be green, then subtracting it from 1.

The probability that the first ball will be not be green is five out of ten, or $\frac{5}{10} = \frac{1}{2}$ .

This will leave nine balls, four not green, making the probability that the second ball will then not be green $\frac{4}{9}$ .

Multiply these probabilities:

$\frac{1}{2} \times \frac{4}{9}= \frac{1}{1} \times \frac{2}{9} = \frac{2}{9}$

This is the probability that neither ball is green; subtract this from 1: $1- \frac{2}{9} = \frac{9}{9}- \frac{2}{9} = \frac{7}{9}$

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Question

Three fair dice are rolled. What is the probability that none of them will show a 5 or a 6?

Answer

The equivalent question is this: What is the probability that all three dice will show 1, 2, 3, or 4?

The probability that one fair die will show a 1, 2, 3, or 4 is $\frac{4}{6} = \frac{2}{3}$ , so the probability that all three will show 1, 2, 3, or 4 is

$\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$ .

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Question

Six boxes are marked with numbers 1 to 6. Each box contains ten marbles, with the number of red marbles in each box corresponding to its number, and the remainder being white.

A fair die is rolled. The box corresponding to the roll is chosen and a marble is drawn at random from that box. What is the probability of drawing a red marble?

Answer

The die is fair, making each box equally likely to be chosen; also, each marble is equally likely to be chosen after the box is chosen. This makes each marble overall equally likely to be chosen. Since there are

red marbles out of

$6 \times 10 = 60$

marbles total, the probability of choosing a red marble is

$\frac{21}{60 } = \frac{21 \div 3}{60 \div 3} = \frac{7}{20}$

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Question

Six boxes are marked with numbers 1 to 6. Each box contains ten marbles, with the number of red marbles in each box corresponding to its number, and the remainder being white.

A fair die is rolled. The box corresponding to the roll is chosen and a marble is drawn at random from that box. What is the probability of drawing a white marble?

Answer

The die is fair, making each box equally likely to be chosen; also, each marble is equally likely to be chosen after the box is chosen. This makes each marble overall equally likely to be chosen. Since there are

white marbles out of

$6 \times 10 = 60$

marbles total, the probability of choosing a white marble is

$\frac{39}{60 } = \frac{39 \div 3}{60 \div 3} = \frac{13}{20}$

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Question

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

Answer

There are 19 winning cards: the 13 clubs, the other 3 queens, and the other 3 kings (the queen and king of clubs are already counted). There are subsequently 33 losing cards, so the odds against a winning card are the number of losing cards, 33, to the number of winning cards, 19. This cannot be reduced, so the correct choice is 33 to 19.

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Question

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?

Answer

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

$\frac{42}{10} = \frac{42 \div 2}{10\div 2} = \frac{21}{5}$

or 21 to 5.

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Question

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?

Answer

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

$\frac{48}{4} = \frac{48 \div 4}{4\div 4} = \frac{12}{1}$

or 12 to 1.

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Question

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?

Answer

Since there is only one card out of 52 (ace of spades) that wins John the \$25 prize, the probability of this happening is $\frac{1}{52}$ . 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 $\frac{3}{52}$ . 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 $\frac{48}{52}$ . The value of this outcome to John is $-\$1$ , 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

$E = \frac{1}{52} \times 24 + \frac{3}{52} \times 4 + \frac{48}{52} \times (-1)$

$E = \frac{24}{52} + \frac{12}{52} - \frac{48}{52} = - \frac{12}{52}$ dollars, or

$-12 \div 52 \approx -\$0.23$

The expected value of the game to Johnny rounds to $-\$0.23$ .

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Question

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?

Answer

Since there is only one card out of 53 (joker) that wins Sharon the \$25 prize, the probability of this happening is $\frac{1}{53}$ . 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 $\frac{1}{53}$ . 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 $\frac{3}{53}$ . 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 $\frac{48}{53}$ . The value of this outcome to Sharon is $-\$2$ , 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

$E = \frac{1}{53} \times 23 + \frac{1}{53} \times 8 + \frac{3}{53} \times 5 + \frac{48}{53} \times (-2)$

$E = \frac{23}{53} + \frac{8}{53} + \frac{15}{53} - \frac{96}{53} = - \frac{50}{53}$ dollars, or

$-50 \div 53 \approx -\$0.94$

The expected value of the game to Sharon rounds to $-\$0.94$ .

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Question

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?

Answer

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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Question

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

Answer

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:

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

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Question

Set A:

Set B:

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

Answer

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:

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

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Question

Set A:

Set B:

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

Answer

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 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:

$P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B)$

The interesection is:

$P(A\bigcap B)=P(A)P(B)=\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

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

$P(A\bigcup B)= \frac{5}{7}+\frac{1}{2}-\frac{5}{14}=\frac{10}{14}+\frac{7}{14}-\frac{5}{14}=\frac{6}{7}$

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Question

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?

Answer

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:

$\frac{2}{42}=\frac{1}{21}$

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Question

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?

Answer

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:

$\frac{200(.20)+(150)(.10)+(250)(.12)+(100)(0.30)}{700}$

$\frac{115}{700}=\frac{23}{140}$

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Question

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 senior student and a student who does not attend honors classes?

Answer

First calculate the number of students:

The percentage of seniors that do not attend honors classes is:

Therefore, the probability of selecting a student who is a senior and one who does not attend honors classes is:

$\frac{100}{700}(.70)=\frac{1}{10}$

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Question

Aaron, Gary, Craig, and Boone are sitting down in a row of four chairs. What is the probability that Aaron and Gary will be seated beside each other?

Answer

Consider first all of the possible ways the men may be arranged, which is

Now, consider all of the ways that Aaron and Gary could be seated beside each other; it may be easier to visualize by drawing it out:

A G _ _

G A _ _

_ A G _

_ G A _

_ _ A G

_ _ G A

As seen, there are six possibilities.

Finally, for each of these cases, Craige and Boone could be seated in one of two ways.

So the probability that Aaron and Gary will be seated beside each other is:

$\frac{2\cdot 6}{24}=\frac{12}{24}=\frac{1}{2}$

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Question

Toss a coin twice. Roll a six sided dice once. What is the probability of flipping two heads and rolling a one?

Answer

Each toss and roll of the dice are independent events. The probability of rolling either a head or tail on the first and second trial is one-half.

The probability of rolling any number on the dice is $\frac{1}{6}$ since there is one number on each of the six faces of the dice.

Multiply all the probabilities.

$P=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{6}= \frac{1}{24}$

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Question

A standard deck of 53 cards (including the joker) is altered by adding the joker and the aces from another deck. What are the odds against a randomly drawn card being an ace?

Answer

Adding the joker and the four aces to a 53-card deck will yield a deck of 58 cards, which contains 8 aces and 50 other cards. The odds against drawing an ace at random are therefore $\frac{50}{8} = \frac{25}{4}$ - that is, 25 to 4.

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