MCQ Test: Theoretical Probability - 1 - Banking Exams MCQ

When drawing without replacement, the probability of the first ball being blue is 8/12. After removing one blue ball, there are 7 blue balls left out of 11 balls in the bag. So, the probability of the second ball being green is 4/11. To find the probability of drawing a blue ball and then a green ball, multiply the individual probabilities: (8/12) * (4/11) = 32/132, which simplifies to 16/88.

The probability of getting a 5 on the first roll of a fair six-sided die is 1/6. The probability of getting an even number on the second roll is 3/6 (there are three even numbers: 2, 4, and 6, out of a total of six possible outcomes). To find the probability of both events occurring, multiply the individual probabilities: (1/6) * (3/6) = 3/36, which simplifies to 1/12.

When drawing without replacement, the probability of the first ball being yellow is 4/10. After removing one yellow ball, there are 3 yellow balls left out of 9 balls in the bag. So, the probability of the second ball being yellow is 3/9. To find the probability of drawing two yellow balls, multiply the individual probabilities: (4/10) * (3/9) = 12/90, which simplifies to 2/15.

When drawing with replacement, the probability of the first ball being black is 3/8, and the probability of the first ball being white is 5/8. Since we are interested in either order (black and then white or white and then black), we need to consider both possibilities. To find the probability of getting one black ball and one white ball in any order, add the individual probabilities: (3/8) * (5/8) + (5/8) * (3/8) = 15/64.

There are 7 red balls and 5 blue balls out of a total of 12 balls (7 red + 5 blue) in the bag. The probability of drawing a red ball or a blue ball is (7 + 5) / 12 = 12/12, which simplifies to 1.

When drawing without replacement, the probability of the first ball being white is 6/10. After removing one white ball, there are 5 white balls left out of 9 balls in the bag. So, the probability of the second ball being white is 5/9. To find the probability of drawing two white balls in a row, multiply the individual probabilities: (6/10) * (5/9) = 30/90, which simplifies to 2/15.

When drawing with replacement, the probability of the first ball being red is 6/14, and the probability of the second ball being red is also 6/14. To find the probability of getting two red balls, multiply the individual probabilities: (6/14) * (6/14) = 36/196, which simplifies to 9/49.

There are two ways to draw two balls of different colors: one yellow and one orange or one orange and one yellow. The probability of drawing one yellow and one orange ball is (5/12) * (7/11) = 35/132. The probability of drawing one orange and one yellow ball is (7/12) * (5/11) = 35/132. Summing up the probabilities of both cases: 35/132 + 35/132 = 70/132, which simplifies to 35/66.

There are two numbers greater than 4 (5 and 6) out of six numbers (1 to 6) on the spinner. The probability of landing on a number greater than 4 is 2/6, which simplifies to 1/3.

When drawing with replacement, the probability of the first ball being black is 8/14, and the probability of the first ball being white is 6/14. Since we are interested in either order (black and then white or white and then black), we need to consider both possibilities. To find the probability of getting a black ball and a white ball in any order, add the individual probabilities: (8/14) * (6/14) + (6/14) * (8/14) = 48/196 + 48/196 = 96/196, which simplifies to 24/49.

When drawing without replacement, the probability of the first ball being red is 6/10. After removing one red ball, there are 5 red balls left out of 9 balls in the box. So, the probability of the second ball being green is 4/9. To find the probability of both events occurring, multiply the individual probabilities: (6/10) * (4/9) = 24/90, which simplifies to 4/15.

There are three possible ways to draw two balls of the same color: two white balls, two red balls, or two blue balls. The probability of drawing two white balls is (4/15) * (3/14) = 12/210. The probability of drawing two red balls is (6/15) * (5/14) = 30/210. The probability of drawing two blue balls is (5/15) * (4/14) = 20/210. Summing up the probabilities of all three cases: 12/210 + 30/210 + 20/210 = 62/210, which simplifies to 31/105.

There are 7 blue balls out of a total of 12 balls (5 red + 7 blue) in the bag. When drawing without replacement, the probability of the first ball being blue is 7/12. After removing one blue ball, there are 6 blue balls left out of 11 balls in the bag. So, the probability of the second ball being blue is 6/11. To find the probability of both events occurring, multiply the individual probabilities: (7/12) * (6/11) = 42/132, which simplifies to 7/132.

When flipping a fair coin, the probability of getting heads on a single flip is 1/2, and the probability of getting tails on a single flip is also 1/2. To get exactly two heads in three flips, there are three possible outcomes: HHT, HTH, and THH, where H represents heads and T represents tails. The probability of each outcome is (1/2) * (1/2) * (1/2) = 1/8. Since there are three favorable outcomes, the total probability of getting exactly two heads is 3/8.

When drawing with replacement, the probability of the first ball being red is 10/18 (10 red balls out of 18 total balls). After replacing the first ball, there are still 10 red balls out of 18 balls in the box. So, the probability of the second ball being red is also 10/18. To find the probability of both events occurring, multiply the individual probabilities: (10/18) * (10/18) = 100/324, which simplifies to 10/36.

There are 4 blue balls and 5 green balls out of a total of 15 balls (6 red + 4 blue + 5 green) in the bag. The probability of drawing a blue or a green ball is (4 + 5) / 15 = 9/15, which simplifies to 3/5.

There are 13 hearts and 13 diamonds out of a total of 52 cards in the deck. The probability of drawing a heart or a diamond is (13 + 13) / 52 = 26/52, which simplifies to 1/2.

There are 5 red balls out of a total of 8 balls (5 red + 3 blue) in the bag. When drawing without replacement, the probability of the first ball being red is 5/8. After removing one red ball, there are 4 red balls left out of 7 balls in the bag. So, the probability of the second ball being red is 4/7. To find the probability of both events occurring, multiply the individual probabilities: (5/8) * (4/7) = 20/56, which simplifies to 5/24.

There are 10 boys out of a total of 25 students in the class. The probability of selecting a boy is 10/25, which simplifies to 2/5.

There are four even numbers (2, 4, 6, 8) out of eight numbers (1 to 8) on the spinner. The probability of landing on an even number is 4/8, which simplifies to 1/2.