This set of Mathematics Assessment Questions for Class 12 focuses on “Multiplication Theorem on Probability”.

1. Which of this represents the multiplication theorem of probability?

a) P(A∩B) = P(B) P(B/A)

b) P(A∩B) = P(A) P(B/A)

c) P(A∩B) = P(A) P(B/B)

d) P(A∩B) = P(A) P(A/A)

View Answer

Explanation: The multiplication theorem of probability is If A and B are two events of a random experiment with P(A) > 0 and P(B) > 0, then P(A∩B) = P(A) P(B/A) or P(A∩B) = P(B) P(A/B).

2. A box contains 5 brown and 7 black pebbles. What is the probability of drawing a brown pebble if the first pebble drawn is black? The balls drawn are not replaced in the box.

a) \(\frac {5}{11}\)

b) \(\frac {8}{11}\)

c) \(\frac {4}{18}\)

d) \(\frac {14}{11}\)

View Answer

Explanation: Total number of balls = 5 + 7 = 12

The probability of the first ball to be blue = \(\frac {7}{12}\)

The probability of drawing red ball as the second ball out without replacement = \(\frac {5}{11}\)

3. Which of this represents the multiplication theorem of probability?

a) P(A∩B) = P(B) P(B/A)

b) P(A∩B) = P(A) P(B/B)

c) P(A∩B) = P(A) P(A/A)

d) P(A∩B) = P(B) P(A/B)

View Answer

Explanation: The multiplication theorem of probability is If A and B are two events of a random experiment with P(A) > 0 and P(B) > 0, then P(A∩B) = P(A) P(B/A) or P(A∩B) = P(B) P(A/B).

4. A bag contains 6 pink and 8 white pebbles. What is the probability of drawing a brown pebble if the first pebble drawn is pink? The balls drawn are not replaced in the bag.

a) 0

b) \(\frac {8}{11}\)

c) 1

d) \(\frac {14}{11}\)

View Answer

Explanation: The total probability is zero because the bag contains only pink and white coloured pebbles. There are no brown coloured pebbles in the bag.

5. A bag contains 9 identical balls, of which are 4 are blue and 6 are green. Three balls are taken out randomly from the bag after one another. Find the probability that all three balls are blue?

a) \(\frac {5}{8}\)

b) \(\frac {6}{19}\)

c) \(\frac {5}{21}\)

d) \(\frac {4}{7}\)

View Answer

Explanation: Since 6 balls are green, the probability of first ball to be drawn is green = \(\frac {6}{9}\)

The probability of the first ball to be green, 8 balls are left and 5 out of them are green = \(\frac {5}{8}\)

If the first two balls are green, then the probability of the third ball to be green = \(\frac {4}{7}\)

According to multiplication theorem, \(\frac {6}{9} . \frac {5}{8} . \frac {4}{7} = \frac {5}{21}\)

6. A box contains 3 red and 4 blue marbles. Two marbles are drawn without replacement. What is the probability that the second marble is red if it is known the first marblel is red?

a) \(\frac {3}{7}\)

b) \(\frac {4}{7}\)

c) \(\frac {1}{3}\)

d) \(\frac {1}{7}\)

View Answer

Explanation: Total number of marbles = 7

Probability of the first marble to be red = \(\frac {3}{7}\)

Probability of the second marble to be red without replacement = \(\frac {2}{6} = \frac {1}{3}\)

7. A bag contains 4 red and 7 blue balls. What is the probability of drawing a blue ball if the first ball drawn is red? The balls drawn are replaced into the bag.

a) \(\frac {8}{11}\)

b) \(\frac {7}{11}\)

c) \(\frac {4}{11}\)

d) \(\frac {7}{4}\)

View Answer

Explanation: Total number of balls = 4 + 7 = 11

The probability of the first ball to be red = \(\frac {4}{11}\)

The probability of drawing blue ball as the second ball out with replacement = \(\frac {7}{11}\)

8. A bag contains 4 red and 7 blue balls. What is the probability of drawing a blue ball if the first ball drawn is red? The balls drawn are not replaced in the bag.

a) \(\frac {7}{10}\)

b) \(\frac {8}{10}\)

c) \(\frac {7}{1}\)

d) \(\frac {4}{11}\)

View Answer

Explanation: Total number of balls = 4 + 7 = 11

The probability of the first ball to be red = \(\frac {4}{11}\)

The probability of drawing blue ball as the second ball out without replacement = \(\frac {7}{10}\)

9. A bag contains 4 red and 7 blue balls. What is the probability of drawing a red ball if the first ball drawn is blue? The balls drawn are not replaced in the bag.

a) \(\frac {7}{11}\)

b) \(\frac {7}{10}\)

c) \(\frac {4}{10}\)

d) \(\frac {9}{11}\)

View Answer

Explanation: Total number of balls = 4 + 7 = 11

The probability of the first ball to be blue = \(\frac {7}{11}\)

The probability of drawing red ball as the second ball out without replacement = \(\frac {4}{10}\)

10. A bag contains 4 red and 7 blue balls. What is the probability of drawing a red ball if the first ball drawn is blue? The balls drawn are replaced in the bag.

a) \(\frac {4}{11}\)

b) \(\frac {8}{11}\)

c) \(\frac {4}{18}\)

d) \(\frac {14}{11}\)

View Answer

Explanation: Total number of balls = 4 + 7 = 11

The probability of the first ball to be blue = \(\frac {7}{11}\)

The probability of drawing red ball as the second ball out without replacement = \(\frac {4}{11}\)

11. A bag contains 3 red, 2 white and 4 green balls. What is the probability of drawing the second ball to be white if the first ball drawn is white? The balls are not replaced in the bag.

a) \(\frac {1}{9}\)

b) \(\frac {2}{9}\)

c) \(\frac {7}{8}\)

d) \(\frac {1}{8}\)

View Answer

Explanation: Total number of balls = 9

The probability of the first ball to be white = \(\frac {2}{9}\)

The probability of drawing white ball again for the second time without replacement = \(\frac {1}{8}\)

12. A bag contains 3 red, 2 white and 4 green balls. What is the probability of drawing the second ball to be white if the first ball drawn is white? The balls are replaced in the bag.

a) \(\frac {1}{9}\)

b) \(\frac {2}{9}\)

c) \(\frac {1}{8}\)

d) \(\frac {2}{8}\)

View Answer

Explanation: Total number of balls = 9

The probability of the first ball to be white = \(\frac {2}{9}\)

The probability of drawing white ball again for the second time without replacement = \(\frac {1}{8}\)

13. A bag contains 3 red, 2 white and 4 green balls. What is the probability of drawing the second ball to be green if the first ball drawn is red? The balls are replaced in the bag.

a) \(\frac {3}{9}\)

b) \(\frac {4}{9}\)

c) \(\frac {4}{3}\)

d) \(\frac {4}{17}\)

View Answer

Explanation: Total number of balls = 9

The probability of the first ball to be red = \(\frac {3}{9}\)

The probability of the second ball to be green with replacement = \(\frac {4}{9}\)

14. A bag contains 3 red, 2 white and 4 green balls. What is the probability of drawing the second ball to be green if the first ball drawn is red? The balls are not replaced in the bag.

a) 0.82

b) 0.91

c) 1.23

d) 0.5

View Answer

Explanation: Total number of balls = 9

The probability of the first ball to be red = \(\frac {3}{9}\)

The probability of the second ball to be green without replacement = \(\frac {4}{8} = \frac {1}{2}\) = 0.5

15. A bag contains 3 red, 2 white and 4 green balls. What is the probability of drawing the second ball to be yellow if the first ball drawn is red? The balls are not replaced in the bag.

a) \(\frac {1}{3}\)

b) \(\frac {1}{2}\)

c) 1

d) 0

View Answer

Explanation: Yellow color ball is not listed in the question. The bag contains only red, white and green color balls. So, the probability to draw a yellow color ball out of the bag is zero.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 12**.

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