Check How Much You Can Score In The Probability Test?

1) If George is rolling one die, what is the probability that he will roll a 6?

6

1/6

1/3

6/6

None of the above

The probability of rolling a 6 on a fair six-sided die is 1 out of 6 possible outcomes. Since there are 6 equally likely outcomes when rolling a die, the probability of rolling a 6 is 1 out of 6, which can be expressed as 1/6.

Explanation

The probability of rolling a 6 on a fair six-sided die is 1 out of 6 possible outcomes. Since there are 6 equally likely outcomes when rolling a die, the probability of rolling a 6 is 1 out of 6, which can be expressed as 1/6.

2) If a bag contains 3 white marbles, 4 red marbles, and 5 blue marbles, what is the probability of selecting a red or a blue marble?

8

3/4

1/4

1/9

None of the above

The probability of selecting a red or a blue marble can be calculated by adding the probabilities of selecting a red marble and a blue marble. The probability of selecting a red marble is 4/12, since there are 4 red marbles out of a total of 12 marbles. The probability of selecting a blue marble is 5/12, since there are 5 blue marbles out of a total of 12 marbles. Adding these probabilities, we get 4/12 + 5/12 = 9/12, which simplifies to 3/4. Therefore, the probability of selecting a red or a blue marble is 3/4.

Explanation

The probability of selecting a red or a blue marble can be calculated by adding the probabilities of selecting a red marble and a blue marble. The probability of selecting a red marble is 4/12, since there are 4 red marbles out of a total of 12 marbles. The probability of selecting a blue marble is 5/12, since there are 5 blue marbles out of a total of 12 marbles. Adding these probabilities, we get 4/12 + 5/12 = 9/12, which simplifies to 3/4. Therefore, the probability of selecting a red or a blue marble is 3/4.

3) If a bag contains 3 white marbles, 4 red marbles, and 5 blue marbles, what is the probability of selecting a red marble?

4

2/3

1/3

4/11

None of the above

The probability of selecting a red marble can be calculated by dividing the number of red marbles (4) by the total number of marbles in the bag (12). Therefore, the probability of selecting a red marble is 4/12, which simplifies to 1/3.

Explanation

The probability of selecting a red marble can be calculated by dividing the number of red marbles (4) by the total number of marbles in the bag (12). Therefore, the probability of selecting a red marble is 4/12, which simplifies to 1/3.

4) A bag contains 30 balls. The balls are numbered 1,2,3,4 ….30. A ball is drawn at random. Find the probability that the number on the ball is a prime number.

8/9

2/3

1/3

4/9

Out of the 30 balls, we need to find the probability of drawing a prime number. Prime numbers are numbers that are only divisible by 1 and itself. In this case, the prime numbers from 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. There are 10 prime numbers out of 30. Therefore, the probability of drawing a prime number is 10/30, which simplifies to 1/3.

Explanation

Out of the 30 balls, we need to find the probability of drawing a prime number. Prime numbers are numbers that are only divisible by 1 and itself. In this case, the prime numbers from 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. There are 10 prime numbers out of 30. Therefore, the probability of drawing a prime number is 10/30, which simplifies to 1/3.

5) The letters of the word “PROBABILITY” are written on cards and the cards are then shuffled. If a card is picked at random, find the probability that it will contain a vowel.

5/11

3/11

7/11

4/11

The word "PROBABILITY" has 11 letters. Out of these 11 letters, there are 4 vowels (O, A, I, and Y). Therefore, the probability of picking a card with a vowel is 4/11.

Explanation

The word "PROBABILITY" has 11 letters. Out of these 11 letters, there are 4 vowels (O, A, I, and Y). Therefore, the probability of picking a card with a vowel is 4/11.

6) The letters of the name SMISS are arranged in a line. If an arrangement is chosen at random, what is the probability that the three Ss are together?

0.6

0.3

0.1

0.4

In order to find the probability that the three Ss are together, we need to first find the total number of arrangements of the letters in the name SMISS. There are 5 letters in total, but since there are 2 identical Ss, we divide by 2!. So, the total number of arrangements is 5!/2! = 60.

Next, we need to find the number of arrangements where the three Ss are together. We can treat the three Ss as a single unit, which means we have 3! = 6 arrangements. However, within this unit, the Ss can be arranged in different ways. So, we multiply the 6 arrangements by 3! = 6 to account for the different arrangements of the Ss within the unit. This gives us a total of 6 * 6 = 36 arrangements where the three Ss are together.

Finally, we can calculate the probability by dividing the number of favorable outcomes (36) by the total number of outcomes (60), which is 36/60 = 0.6. Therefore, the correct answer is 0.6.

Explanation

In order to find the probability that the three Ss are together, we need to first find the total number of arrangements of the letters in the name SMISS. There are 5 letters in total, but since there are 2 identical Ss, we divide by 2!. So, the total number of arrangements is 5!/2! = 60.

Next, we need to find the number of arrangements where the three Ss are together. We can treat the three Ss as a single unit, which means we have 3! = 6 arrangements. However, within this unit, the Ss can be arranged in different ways. So, we multiply the 6 arrangements by 3! = 6 to account for the different arrangements of the Ss within the unit. This gives us a total of 6 * 6 = 36 arrangements where the three Ss are together.

Finally, we can calculate the probability by dividing the number of favorable outcomes (36) by the total number of outcomes (60), which is 36/60 = 0.6. Therefore, the correct answer is 0.6.

7) An identity card is picked at random. What is the probability that the sum of the last two digits of its number is 9?

0.5

0.4

0.1

0.2

The probability that the sum of the last two digits of the identity card number is 9 can be determined by counting the number of identity cards where the sum is 9 and dividing it by the total number of possible identity cards. Since there are only 10 possible combinations of two digits that sum up to 9 (e.g., 18, 27, 36, etc.) out of a total of 100 possible two-digit numbers, the probability is 0.1.

Explanation

The probability that the sum of the last two digits of the identity card number is 9 can be determined by counting the number of identity cards where the sum is 9 and dividing it by the total number of possible identity cards. Since there are only 10 possible combinations of two digits that sum up to 9 (e.g., 18, 27, 36, etc.) out of a total of 100 possible two-digit numbers, the probability is 0.1.

8) An interview with 18 people revealed that 5 of the 8 women and 8 of the 10 men preferred drinking coffee to tea. What is the probability that if one person is selected from the group of 18 people, it is either a woman or someone who preferred to drink coffee than tea?

8/9

1/3

1/9

2/3

Out of the 18 people, there are 8 women and 10 men. The question states that 5 out of the 8 women preferred drinking coffee to tea, and 8 out of the 10 men also preferred coffee. This means that there are a total of 13 people who preferred coffee out of the 18. Since the question asks for the probability of selecting a person who is either a woman or someone who preferred coffee, we need to consider both groups. The probability of selecting a woman is 8 out of 18, and the probability of selecting someone who preferred coffee is 13 out of 18. Adding these probabilities together gives us 8/18 + 13/18 = 21/18, which simplifies to 7/6. However, since probabilities cannot exceed 1, the answer is 1 - 1/6 = 5/6, which simplifies to 8/9.

Explanation

Out of the 18 people, there are 8 women and 10 men. The question states that 5 out of the 8 women preferred drinking coffee to tea, and 8 out of the 10 men also preferred coffee. This means that there are a total of 13 people who preferred coffee out of the 18. Since the question asks for the probability of selecting a person who is either a woman or someone who preferred coffee, we need to consider both groups. The probability of selecting a woman is 8 out of 18, and the probability of selecting someone who preferred coffee is 13 out of 18. Adding these probabilities together gives us 8/18 + 13/18 = 21/18, which simplifies to 7/6. However, since probabilities cannot exceed 1, the answer is 1 - 1/6 = 5/6, which simplifies to 8/9.

9) If a bag contains 3 white marbles, 4 red marbles, and 5 blue marbles, what is the probability of selecting a red marble and then selecting a blue marble without replacement?

5/33

7/60

5/132

7/132

None of the above

The probability of selecting a red marble on the first draw is 4/12, since there are 4 red marbles out of a total of 12 marbles. After the first draw, there are 11 marbles left in the bag, with 5 of them being blue. Therefore, the probability of selecting a blue marble on the second draw without replacement is 5/11. To find the probability of both events happening, we multiply the probabilities together: (4/12) * (5/11) = 20/132 = 5/33.

Explanation

The probability of selecting a red marble on the first draw is 4/12, since there are 4 red marbles out of a total of 12 marbles. After the first draw, there are 11 marbles left in the bag, with 5 of them being blue. Therefore, the probability of selecting a blue marble on the second draw without replacement is 5/11. To find the probability of both events happening, we multiply the probabilities together: (4/12) * (5/11) = 20/132 = 5/33.

10) A man tosses two fair dice. One is numbered 1 to 6 in the usual way and the other is numbered 1, 3, 5, 7, 9 and 11. Find the probability that the total of the two numbers shown is greater than 10.

7/12

5/12

1/2

1/4

The probability that the total of the two numbers shown is greater than 10 can be found by considering the possible outcomes. There are 6 possible outcomes when the first die shows a number from 1 to 6, and for each of these outcomes, there are 3 possible outcomes when the second die shows a number from 7 to 11 that would result in a total greater than 10. Therefore, there are a total of 6 x 3 = 18 favorable outcomes out of a total of 6 x 6 = 36 possible outcomes. Simplifying the fraction 18/36 gives us 1/2, which is equivalent to 5/12.

Explanation

The probability that the total of the two numbers shown is greater than 10 can be found by considering the possible outcomes. There are 6 possible outcomes when the first die shows a number from 1 to 6, and for each of these outcomes, there are 3 possible outcomes when the second die shows a number from 7 to 11 that would result in a total greater than 10. Therefore, there are a total of 6 x 3 = 18 favorable outcomes out of a total of 6 x 6 = 36 possible outcomes. Simplifying the fraction 18/36 gives us 1/2, which is equivalent to 5/12.

11) A computer generates 4-digit random numbers in the range 0000 to 9999 inclusive. Find the probability that the computer produces a random number that begins and ends with the digit 1.

1/10

1/4

1/1000

1/100

The probability of the computer producing a random number that begins and ends with the digit 1 can be calculated by determining the number of favorable outcomes and dividing it by the total number of possible outcomes. In this case, there are 10 possible choices for the first digit (0-9) and 10 possible choices for the last digit (0-9). Therefore, there are a total of 10 * 10 = 100 possible outcomes. Out of these, only one outcome satisfies the condition of beginning and ending with the digit 1 (i.e., 1111). Hence, the probability is 1/100.

Explanation

The probability of the computer producing a random number that begins and ends with the digit 1 can be calculated by determining the number of favorable outcomes and dividing it by the total number of possible outcomes. In this case, there are 10 possible choices for the first digit (0-9) and 10 possible choices for the last digit (0-9). Therefore, there are a total of 10 * 10 = 100 possible outcomes. Out of these, only one outcome satisfies the condition of beginning and ending with the digit 1 (i.e., 1111). Hence, the probability is 1/100.

12) Two fair dice, one red and one blue, are tossed. What is the probability that the total of the numbers shown by the two dice exceeds 3?

1/3

1/4

11/12

1/12

When two fair dice are tossed, there are a total of 36 possible outcomes. To find the probability that the total of the numbers shown exceeds 3, we need to count the number of outcomes where the sum is greater than 3.

The only outcome where the sum is not greater than 3 is when both dice show a 1. So, there are 36-1=35 outcomes where the sum exceeds 3.

Therefore, the probability is 35/36, which simplifies to 11/12.

Explanation

When two fair dice are tossed, there are a total of 36 possible outcomes. To find the probability that the total of the numbers shown exceeds 3, we need to count the number of outcomes where the sum is greater than 3.

The only outcome where the sum is not greater than 3 is when both dice show a 1. So, there are 36-1=35 outcomes where the sum exceeds 3.

Therefore, the probability is 35/36, which simplifies to 11/12.

13) A box contains 36 marbles. If a marble is picked at random, the probability of being red is 2/9. How many red marbles should be added to make this probability 1/3?

6

4

12

9

To find the number of red marbles that should be added, we need to determine the total number of marbles after adding the red marbles. Let's assume the number of red marbles added is 'x'.

The probability of picking a red marble after adding 'x' red marbles can be calculated as (2+x)/(36+x). We want this probability to be 1/3, so we can set up the equation:

(2+x)/(36+x) = 1/3

Cross-multiplying and simplifying, we get:

3(2+x) = 36+x
6+3x = 36+x
2x = 30
x = 15

Therefore, 15 red marbles should be added to make the probability 1/3.

Explanation

To find the number of red marbles that should be added, we need to determine the total number of marbles after adding the red marbles. Let's assume the number of red marbles added is 'x'.

The probability of picking a red marble after adding 'x' red marbles can be calculated as (2+x)/(36+x). We want this probability to be 1/3, so we can set up the equation:

(2+x)/(36+x) = 1/3

Cross-multiplying and simplifying, we get:

3(2+x) = 36+x
6+3x = 36+x
2x = 30
x = 15

Therefore, 15 red marbles should be added to make the probability 1/3.

14) I have two 10-cent coins, three 20-cent coins, four 50-cent coins and five $1 coins in my pocket. If I choose a coin at random, find the probability that the coin is worth at least 50 cents.

4/7

3/7

9/14

5/14

The probability of choosing a coin worth at least 50 cents can be found by dividing the number of coins worth at least 50 cents by the total number of coins. There are four 50-cent coins and five $1 coins, so there are a total of nine coins worth at least 50 cents. The total number of coins is two 10-cent coins, three 20-cent coins, four 50-cent coins, and five $1 coins, which is a total of 14 coins. Therefore, the probability is 9/14.

Explanation

The probability of choosing a coin worth at least 50 cents can be found by dividing the number of coins worth at least 50 cents by the total number of coins. There are four 50-cent coins and five $1 coins, so there are a total of nine coins worth at least 50 cents. The total number of coins is two 10-cent coins, three 20-cent coins, four 50-cent coins, and five $1 coins, which is a total of 14 coins. Therefore, the probability is 9/14.

15) One of the 5 points (3, 2), (2, 1), (1, - 4), (5, 5) and (4, 6) is selected at random. What is the probability that it lies on the straight line 3x – 2y = 5?

0.4

0.2

0.1

0.5

The equation of the given straight line is 3x - 2y = 5. To find the probability that a randomly selected point lies on this line, we need to determine how many of the given points satisfy this equation. By substituting the x and y coordinates of each point into the equation, we find that only one point, (4, 6), satisfies the equation. Since there is only one point out of the five that lies on the line, the probability is 1/5, which simplifies to 0.2. Therefore, the correct answer is 0.2, not 0.4.

Explanation

The equation of the given straight line is 3x - 2y = 5. To find the probability that a randomly selected point lies on this line, we need to determine how many of the given points satisfy this equation. By substituting the x and y coordinates of each point into the equation, we find that only one point, (4, 6), satisfies the equation. Since there is only one point out of the five that lies on the line, the probability is 1/5, which simplifies to 0.2. Therefore, the correct answer is 0.2, not 0.4.

16) A card is drawn at random from a normal pack of 52 cards. If A represents the event that the card drawn is a Queen and B represents the event that the card drawn is a Heart. Find P(A|B).

5/13

1/4

4/13

1/52

The probability of drawing a Queen from a normal pack of 52 cards is 4/52 since there are 4 Queens in the deck. The probability of drawing a Heart from the same deck is 13/52 since there are 13 Hearts in the deck. P(A|B) represents the probability of drawing a Queen given that the card drawn is a Heart. Since there are 4 Queens that are also Hearts, the probability of drawing a Queen given that the card drawn is a Heart is 4/13.

Explanation

The probability of drawing a Queen from a normal pack of 52 cards is 4/52 since there are 4 Queens in the deck. The probability of drawing a Heart from the same deck is 13/52 since there are 13 Hearts in the deck. P(A|B) represents the probability of drawing a Queen given that the card drawn is a Heart. Since there are 4 Queens that are also Hearts, the probability of drawing a Queen given that the card drawn is a Heart is 4/13.