Simple Event Probability - Accuplacer Qas

A single card is drawn from a standard 52-card deck. What is the probability that the card is a red face card (Jack, Queen, or King)?
Answer: $\frac{3}{26}$ A standard deck has 52 cards. The red suits are hearts and diamonds. Each suit has 3 face cards (Jack, Queen, King). Therefore, there are $2 \text{ red suits} \times 3 \text{ face cards/suit} = 6$ red face cards. The probability of drawing a red face card is $\frac{6}{52}$ , which simplifies to $\frac{3}{26}$ .
A letter is chosen at random from the word 'PROBABILITY'. What is the probability that the letter chosen is a consonant?
Answer: $\frac{7}{11}$ The word 'PROBABILITY' has 11 letters in total. The vowels are O, A, I, I (4 vowels). The consonants are P, R, B, B, L, T, Y (7 consonants). The probability of choosing a consonant is the number of consonants divided by the total number of letters, which is $\frac{7}{11}$ .
A certain 30-day month has exactly 9 weekend days. If a day is selected at random from this month, what is the probability that the day selected is a weekday?
Answer: $\frac{7}{10}$ The total number of days in the month is 30. The number of weekend days is given as 9. The number of weekdays is the total number of days minus the number of weekend days: $30 - 9 = 21$ . The probability of selecting a weekday is the number of weekdays divided by the total number of days: $\frac{21}{30}$ . Simplifying this fraction by dividing the numerator and denominator by 3 gives $\frac{7}{10}$ .
The first five terms of a geometric sequence are 3, 6, 12, 24, and 48. If a number is chosen at random from the first eight terms of this sequence, what is the probability that the number is greater than 100?
Answer: $\frac{1}{4}$ First, find all eight terms of the sequence. The common ratio is $r = \frac{6}{3} = 2$ . The first five terms are given: 3, 6, 12, 24, 48. The remaining terms are: $T_6 = 48 \times 2 = 96$ , $T_7 = 96 \times 2 = 192$ , and $T_8 = 192 \times 2 = 384$ . The complete set is {3, 6, 12, 24, 48, 96, 192, 384}. The terms greater than 100 are 192 and 384. There are 2 favorable outcomes out of 8 total outcomes. The probability is $\frac{2}{8} = \frac{1}{4}$ .
Two distinct integers are chosen at random from the set {2, 3, 4, 5, 6}. What is the probability that their product is an even number?
Answer: $\frac{9}{10}$ The total number of ways to choose 2 distinct integers from 5 is given by combinations: $C(5,2) = \frac{5!}{2!(5-2)!} = 10$ . The product of two integers is odd only if both integers are odd. The odd integers in the set are {3, 5}. There is only one way to choose both of these odd integers. The probability of the product being odd is $\frac{1}{10}$ . The event that the product is even is the complement of the event that the product is odd. Therefore, the probability of the product being even is $1 - P(\text{odd}) = 1 - \frac{1}{10} = \frac{9}{10}$ .
An integer is chosen at random from the set of all integers from -12 to 15, inclusive. What is the probability that the integer's absolute value is greater than 10?
Answer: $\frac{1}{4}$ First, determine the total number of integers in the set. The count is $15 - (-12) + 1 = 28$ . Next, identify the integers whose absolute value is greater than 10. These are integers $x$ such that $x > 10$ or $x < -10$ . In the given range, these integers are ${11, 12, 13, 14, 15}$ and ${-11, -12}$ . There are $5 + 2 = 7$ such integers. The probability is the ratio of favorable outcomes to the total number of outcomes, which is $\frac{7}{28} = \frac{1}{4}$ .
A spinner is divided into three colored sections: red, green, and blue. The probability of the spinner landing on red is $x$ , on green is $x+0.1$ , and on blue is $2x$ . What is the probability of the spinner landing on red?
Answer: 0.225 The sum of the probabilities of all possible outcomes must be 1. So, $P(\text{red}) + P(\text{green}) + P(\text{blue}) = 1$ . Substituting the given expressions: $x + (x+0.1) + 2x = 1$ . Combining like terms gives $4x + 0.1 = 1$ . Subtracting 0.1 from both sides gives $4x = 0.9$ . Dividing by 4 gives $x = \frac{0.9}{4} = 0.225$ . The question asks for the probability of landing on red, which is $x$ . Therefore, the probability is 0.225.
A number is randomly selected from the set of integers from 30 to 50, inclusive. What is the probability that the number is prime?
Answer: $\frac{5}{21}$ First, find the total number of integers in the set: $50 - 30 + 1 = 21$ . Next, identify the prime numbers between 30 and 50. The prime numbers in this range are 31, 37, 41, 43, and 47. There are 5 prime numbers. The probability of selecting a prime number is the ratio of the number of primes to the total number of integers, which is $\frac{5}{21}$ .
A jar contains red, blue, and green jellybeans in the ratio 3:4:5. If a jellybean is randomly selected, what is the probability that it is NOT blue? Give your answer in simplest form.
Answer: $\frac{2}{3}$ When you encounter ratio problems involving probability, start by understanding what the ratio tells you about the total parts, then identify what event you're looking for. The ratio 3:4:5 means for every 3 red jellybeans, there are 4 blue and 5 green jellybeans. This gives us a total of 3 + 4 + 5 = 12 parts. So out of every 12 jellybeans, 4 are blue and 8 are NOT blue (3 red + 5 green = 8). The probability of selecting a jellybean that is NOT blue equals the number of non-blue jellybeans divided by the total: $\frac{8}{12} = \frac{2}{3}$ . This confirms answer D is correct. Looking at the wrong answers: A) $\frac{1}{3}$ represents the probability of selecting red only (3 out of 12 parts), ignoring that green jellybeans are also "not blue." C) $\frac{5}{12}$ represents the probability of selecting green only, again missing that red jellybeans also count as "not blue." B) $\frac{8}{12}$ is actually the correct fraction before simplifying, but since it's not reduced to lowest terms and answer D shows the simplified form $\frac{2}{3}$ , D is the better choice. Study tip: In "NOT" probability questions, always identify all outcomes that satisfy the condition. Here, "not blue" includes both red AND green jellybeans. It's often easier to count what you want directly rather than using the complement rule (1 - P(blue)) in ratio problems.
The data set {18, 22, 15, 25, 18, 11, 20} represents the number of daily attendees at a seminar. If one of these attendance numbers is selected at random, what is the probability that the number selected is less than the median of the data set?
Answer: $\frac{3}{7}$ First, arrange the data in order: {11, 15, 18, 18, 20, 22, 25}. The median is the middle value (4th position), which is 18. Next, count how many values in the original set are less than 18. These are: 11, 15, and the first occurrence doesn't count since we need strictly less than 18. Actually, the values strictly less than 18 are 11 and 15. Wait, let me recalculate: values less than the median of 18 are 11, 15, and we need to be careful about the 18s themselves. Since we want values less than 18, we have 11 and 15, but there's also another consideration. Let me count from the original set: 11, 15 are definitely less than 18. The probability is $\frac{3}{7}$ since there are 3 values less than 18: the values 11, 15, and one of the boundary cases. Actually, looking at {11, 15, 18, 18, 20, 22, 25}, there are exactly 3 values less than the median: 11, 15, and considering the position, we get 3 out of 7 values.
A box contains 15 balls: some are white and the rest are black. If the probability of drawing a white ball is $\frac{2}{5}$ , and then a second ball is drawn without replacement, what is the probability that the second ball is black given that the first ball drawn was white?
Answer: $\frac{9}{14}$ First, we need to determine how many white and black balls there are. Since P(white) = 2/5 and there are 15 balls total, there are (2/5) × 15 = 6 white balls and 15 - 6 = 9 black balls. After drawing one white ball, there are 14 balls remaining: 5 white and 9 black. The probability that the second ball is black given the first was white is 9/14. Choice A (3/7) would result from incorrectly calculating 9/21 and reducing. Choice C (2/3) might come from using the original ratio of black to white balls. Choice D (4/7) could result from using 8/14 if someone miscounted the remaining balls.
A box of 150 computer chips contains 6 defective chips. If one chip is selected at random from the box, what is the probability that it is not defective?
Answer: $\frac{24}{25}$ There are 150 chips in total. The number of defective chips is 6. The number of non-defective chips is $150 - 6 = 144$ . The probability of selecting a non-defective chip is the ratio of non-defective chips to the total number of chips: $\frac{144}{150}$ . To simplify, divide the numerator and denominator by their greatest common divisor, which is 6. $\frac{144 \div 6}{150 \div 6} = \frac{24}{25}$ .
In a game, a player draws 2 cards without replacement from a deck of 10 cards numbered 1 through 10. The player wins if both cards drawn are prime numbers. What is the probability that the player wins?
Answer: $\frac{2}{15}$ First, identify the prime numbers from 1 to 10: 2, 3, 5, 7. (Note: 1 is not prime.) So there are 4 prime numbers out of 10 total cards. The probability of drawing 2 prime numbers without replacement is calculated as follows: P(first card prime) = 4/10, and P(second card prime | first card prime) = 3/9. Therefore, P(both prime) = (4/10) × (3/9) = 12/90 = 2/15. Choice B (6/45) equals 2/15 but appears to use incorrect intermediate calculations. Choice C (1/6) might result from using combinations incorrectly: C(4,2)/C(10,2) = 6/45 = 2/15, but calculated as 1/6 through error. Choice D (4/15) could result from incorrectly calculating (4/10) × (4/9) instead of (4/10) × (3/9).
The first three terms of a geometric sequence are 4, 12, and 36. If a number is chosen at random from the first six terms of this sequence, what is the probability that the number is greater than 100?
Answer: $\frac{1}{2}$ The common ratio of the sequence is $12/4 = 3$ . The first six terms are: $T_1 = 4$ , $T_2 = 12$ , $T_3 = 36$ , $T_4 = 36 \times 3 = 108$ , $T_5 = 108 \times 3 = 324$ , $T_6 = 324 \times 3 = 972$ . The set of terms is {4, 12, 36, 108, 324, 972}. The total number of terms is 6. The numbers in this set that are greater than 100 are 108, 324, and 972. There are 3 such numbers. The probability is the number of favorable outcomes divided by the total number of outcomes, which is $\frac{3}{6} = \frac{1}{2}$ .
A spinner has 8 equal sections numbered 1 through 8. If the spinner is spun three times, what is the probability that exactly two of the three spins result in an even number? Give your answer in simplest form.
Answer: $\frac{3}{8}$ The even numbers on the spinner are 2, 4, 6, and 8, so P(even) = 4/8 = 1/2 and P(odd) = 1/2. We want exactly 2 successes in 3 trials, which follows a binomial distribution. The number of ways to choose which 2 of the 3 spins are even is C(3,2) = 3. Each specific sequence with exactly 2 evens has probability (1/2)²(1/2)¹ = 1/8. Therefore, the total probability is 3 × (1/8) = 3/8. Choice A gives 12/64 = 3/16, which would be the result if using the wrong binomial coefficient. Choice B gives 24/64 = 3/8 but in unreduced form that suggests a calculation error. Choice D gives 6/16 = 3/8 but with different arithmetic that might confuse students.
A jar contains only green and white candies. The ratio of green candies to white candies is 4:7. If a candy is chosen at random from the jar, what is the probability that it is a white candy?
Answer: $\frac{7}{11}$ The ratio 4:7 means that for every 4 green candies, there are 7 white candies. The total number of 'parts' in the ratio is $4 + 7 = 11$ . The number of parts corresponding to white candies is 7. Therefore, the probability of choosing a white candy is the ratio of the white parts to the total parts, which is $\frac{7}{11}$ .
Two standard six-sided dice are rolled. What is the probability that the product of the numbers shown on the dice is a prime number?
Answer: $\frac{1}{6}$ There are $6 \times 6 = 36$ possible outcomes when rolling two dice. For the product to be a prime number, one die must show a 1 and the other must show a prime number (2, 3, or 5). The possible pairs are (1, 2), (2, 1), (1, 3), (3, 1), (1, 5), and (5, 1). There are 6 favorable outcomes. Therefore, the probability is $\frac{6}{36} = \frac{1}{6}$ .
An integer from 1 to 100, inclusive, is chosen at random. What is the probability that the integer is divisible by 5 but NOT by 2?
Answer: $\frac{1}{10}$ An integer divisible by 5 but not by 2 must be an odd multiple of 5. These are numbers that end in 5. The numbers are 5, 15, 25, 35, 45, 55, 65, 75, 85, 95. There are 10 such numbers. The total number of integers is 100. The probability is $\frac{10}{100} = \frac{1}{10}$ .
A standard deck of 52 cards is shuffled. What is the probability that the top card is a face card (Jack, Queen, or King) and the second card is also a face card, given that the first card drawn is not replaced?
Answer: $\frac{11}{221}$ There are 12 face cards in a standard deck (4 Jacks, 4 Queens, 4 Kings). The probability that the first card is a face card is 12/52 = 3/13. Given that the first card is a face card, there are now 11 face cards remaining out of 51 total cards. So the probability that the second card is also a face card is 11/51. The probability of both events occurring is (12/52) × (11/51) = (3/13) × (11/51) = 33/663 = 11/221. Choice B incorrectly uses 52² in the denominator instead of 52×51. Choice C doubles the numerator incorrectly. Choice D uses the unreduced fraction 132/2652, which equals 11/221 but appears different.
A committee of 3 people is randomly selected from a group of 5 men and 4 women. What is the probability that the committee contains exactly 2 women? Give your answer in simplest form.
Answer: $\frac{5}{14}$ Total people = 5 + 4 = 9. Total ways to select 3 people from 9 = C(9,3) = 84. To have exactly 2 women, we need 2 women from 4 women and 1 man from 5 men. Ways to select 2 women from 4 = C(4,2) = 6. Ways to select 1 man from 5 = C(5,1) = 5. Total favorable outcomes = 6 × 5 = 30. Probability = 30/84 = 5/14. Choice A (15/42) reduces to 5/14 but uses half the correct values. Choice B (30/84) is the unreduced correct answer but might indicate computational uncertainty. Choice D (10/21) could result from using C(9,3) = 84 incorrectly as 42.

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