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Math 110 Module 5: Probability And Counting Rules Exam 2026/2027 Update. Just Released

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MATH 110 MODULE 5: PROBABILITY AND COUNTING RULES EXAM 2026/2027 UPDATE. JUST RELEASED

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MATH 110 MODULE 5: PROBABILITY
AND COUNTING RULES EXAM 2026/2027
UPDATE. JUST RELEASED


1. A jar contains 5 red marbles, 3 blue marbles, and 2 green marbles. If two marbles are

drawn without replacement, what is the probability that both are red?

A. 1/4


B. 2/9


C. 5/18


D. 25/100


Answer: B


Conceptual Explanation: The probability of the first red marble is 5/10. Since it is without

replacement, the probability of the second red marble is 4/9. (5/10) * (4/9) = 20/90 = 2/9.


2. Which of the following conditions must be met for two events A and B to be considered

mutually exclusive?

A. P(A and B) = 0


B. P(A or B) = P(A) + P(B) - P(A and B)


C. P(A and B) = P(A) * P(B)


D. P(A | B) = P(A)

,Answer: A


Conceptual Explanation: Events are mutually exclusive if they cannot occur at the same

time, meaning their intersection (A and B) is zero.


3. If the probability of an event occurring is 0.35, what are the odds against the event?

A. 7:13


B. 13:7


C. 7:20


D. 20:7


Answer: B


Conceptual Explanation: P(event) = 0.35 = 7/20. P(not event) = 0.65 = 13/20. Odds

against are P(not event) : P(event), which is 13:7.


4. How many different 4-letter permutations can be formed from the letters in the word

‘NUMBER’?

A. 360


B. 720


C. 24


D. 15


Answer: A

, Conceptual Explanation: The word ‘NUMBER’ has 6 distinct letters. The number of 4-

letter permutations is 6P4 = 6 * 5 * 4 * 3 = 360.


5. A committee of 3 people is to be formed from a group of 8. How many different

committees are possible?

A. 56


B. 336


C. 24


D. 112


Answer: A


Conceptual Explanation: Since the order does not matter in a committee, use

combinations: 8C3 = (8 * 7 * 6) / (3 * 2 * 1) = 56.


6. If P(A) = 0.6, P(B) = 0.3, and A and B are independent, what is P(A or B)?

A. 0.72


B. 0.18


C. 0.90


D. 0.42


Answer: A


Conceptual Explanation: For independent events, P(A and B) = P(A)P(B) = 0.60.3 = 0.18.

P(A or B) = P(A) + P(B) - P(A and B) = 0.6 + 0.3 - 0.18 = 0.72.

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