Probability: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 1)

P(A union B) is the probability that A occurs, B occurs, or both occur. Simply adding P(A) and P(B) counts the outcomes common to both events twice. So P(A intersection B) is subtracted once to correct this double-count. This gives P(A union B) = P(A) + P(B) – P(A intersection B).

A standard pack of playing cards has 52 cards in total. These cards are divided into 4 suits: spades, clubs, hearts, and diamonds. Each suit has 13 cards, since 4 x 13 = 52. Spades and clubs are black, while hearts and diamonds are red, giving 26 cards of each colour.

The favourable count n(E) can never be fewer than 0 or more than n(S). So the ratio n(E)/n(S) always lies between 0 and 1. A probability of 0 means the event is impossible, and 1 means it is certain. So for any event E, 0 <= P(E) <= 1.

An impossible event is one that can never occur in the experiment. It has no favourable outcomes, so n(E) = 0. Using P(E) = n(E)/n(S), this gives P(E) = 0/n(S) = 0. So every impossible event has a probability of 0.

The sample space S contains every possible outcome of the experiment. So whenever the experiment is performed, some outcome in S is certain to occur. An event that is certain to happen has a probability of 1. So P(S) = 1.

Let n(S) be the total number of possible outcomes in the sample space. Let n(E) be the number of outcomes favourable to event E. The probability of E is then defined as P(E) = n(E) / n(S). This formula assumes every outcome in S is equally likely.

A random experiment can have several different results. The set of all these possible results is called the sample space. It is usually denoted by the letter S. For example, tossing a coin once has sample space S = {H, T}.

We are told P(E) = 1/4. The events E and ‘not E’ are complementary, so together they cover the entire sample space. This means P(E) + P(not E) = 1. So P(not E) = 1 – P(E) = 1 – 1/4 = 3/4. Check: P(E) + P(not E) = 1/4 + 3/4 = 1.

We are told P(E) = 3/10. The events E and ‘not E’ are complementary, so together they cover the entire sample space. This means P(E) + P(not E) = 1. So P(not E) = 1 – P(E) = 1 – 3/10 = 7/10. Check: P(E) + P(not E) = 3/10 + 7/10 = 1.

We are told P(E) = 1/6. The events E and ‘not E’ are complementary, so together they cover the entire sample space. This means P(E) + P(not E) = 1. So P(not E) = 1 – P(E) = 1 – 1/6 = 5/6. Check: P(E) + P(not E) = 1/6 + 5/6 = 1.

We are told P(E) = 7/20. The events E and ‘not E’ are complementary, so together they cover the entire sample space. This means P(E) + P(not E) = 1. So P(not E) = 1 – P(E) = 1 – 7/20 = 13/20. Check: P(E) + P(not E) = 7/20 + 13/20 = 1.

We are told P(E) = 1/3. The events E and ‘not E’ are complementary, so together they cover the entire sample space. This means P(E) + P(not E) = 1. So P(not E) = 1 – P(E) = 1 – 1/3 = 2/3. Check: P(E) + P(not E) = 1/3 + 2/3 = 1.

We are told P(E) = 5/12. The events E and ‘not E’ are complementary, so together they cover the entire sample space. This means P(E) + P(not E) = 1. So P(not E) = 1 – P(E) = 1 – 5/12 = 7/12. Check: P(E) + P(not E) = 5/12 + 7/12 = 1.

We are told P(E) = 2/5. The events E and ‘not E’ are complementary, so together they cover the entire sample space. This means P(E) + P(not E) = 1. So P(not E) = 1 – P(E) = 1 – 2/5 = 3/5. Check: P(E) + P(not E) = 2/5 + 3/5 = 1.

We are told P(E) = 3/8. The events E and ‘not E’ are complementary, so together they cover the entire sample space. This means P(E) + P(not E) = 1. So P(not E) = 1 – P(E) = 1 – 3/8 = 5/8. Check: P(E) + P(not E) = 3/8 + 5/8 = 1.

The bag has 8 red balls, 9 blue balls, and 6 green balls, which is 23 balls in total. We draw 2 balls at random, so n(S) = 23C2 = 253, the number of ways to choose any 2 balls. We want both balls to be blue, and there are 9 blue balls in the bag. The number of ways to pick 2 blue balls is 9C2 = 36. So n(E) = 36. Using P(E) = n(E)/n(S), P(E) = 36/253 = 36/253.

The bag has 10 red balls, 3 blue balls, and 6 green balls, which is 19 balls in total. We draw 2 balls at random, so n(S) = 19C2 = 171, the number of ways to choose any 2 balls. We want both balls to be green, and there are 6 green balls in the bag. The number of ways to pick 2 green balls is 6C2 = 15. So n(E) = 15. Using P(E) = n(E)/n(S), P(E) = 15/171 = 5/57.

The bag has 5 red balls, 6 blue balls, and 1 green ball, which is 12 balls in total. We draw 2 balls at random, so n(S) = 12C2 = 66, the number of ways to choose any 2 balls. We want both balls to be blue, and there are 6 blue balls in the bag. The number of ways to pick 2 blue balls is 6C2 = 15. So n(E) = 15. Using P(E) = n(E)/n(S), P(E) = 15/66 = 5/22.

The bag has 5 red balls, 9 blue balls, which is 14 balls in total. We draw 2 balls at random, so n(S) = 14C2 = 91, the number of ways to choose any 2 balls. We want both balls to be blue, and there are 9 blue balls in the bag. The number of ways to pick 2 blue balls is 9C2 = 36. So n(E) = 36. Using P(E) = n(E)/n(S), P(E) = 36/91 = 36/91.

The bag has 4 red balls, 3 blue balls, and 2 green balls, which is 9 balls in total. We draw 2 balls at random, so n(S) = 9C2 = 36, the number of ways to choose any 2 balls. We want both balls to be blue, and there are 3 blue balls in the bag. The number of ways to pick 2 blue balls is 3C2 = 3. So n(E) = 3. Using P(E) = n(E)/n(S), P(E) = 3/36 = 1/12.

The bag has 7 red balls, 6 blue balls, and 1 green ball, which is 14 balls in total. We draw 2 balls at random, so n(S) = 14C2 = 91, the number of ways to choose any 2 balls. We want both balls to be blue, and there are 6 blue balls in the bag. The number of ways to pick 2 blue balls is 6C2 = 15. So n(E) = 15. Using P(E) = n(E)/n(S), P(E) = 15/91 = 15/91.

The bag has 6 red balls, 3 blue balls, and 6 green balls, which is 15 balls in total. We draw 2 balls at random, so n(S) = 15C2 = 105, the number of ways to choose any 2 balls. We want both balls to be green, and there are 6 green balls in the bag. The number of ways to pick 2 green balls is 6C2 = 15. So n(E) = 15. Using P(E) = n(E)/n(S), P(E) = 15/105 = 1/7.

The bag has 3 red balls, 9 blue balls, and 4 green balls, which is 16 balls in total. We draw 2 balls at random, so n(S) = 16C2 = 120, the number of ways to choose any 2 balls. We want both balls to be blue, and there are 9 blue balls in the bag. The number of ways to pick 2 blue balls is 9C2 = 36. So n(E) = 36. Using P(E) = n(E)/n(S), P(E) = 36/120 = 3/10.

The bag has 10 red balls, 3 blue balls, and 6 green balls, which is 19 balls in total. We draw 2 balls at random, so n(S) = 19C2 = 171, the number of ways to choose any 2 balls. We want both balls to be red, and there are 10 red balls in the bag. The number of ways to pick 2 red balls is 10C2 = 45. So n(E) = 45. Using P(E) = n(E)/n(S), P(E) = 45/171 = 5/19.

The bag has 7 red balls, 8 blue balls, and 3 green balls, which is 18 balls in total. We draw 2 balls at random, so n(S) = 18C2 = 153, the number of ways to choose any 2 balls. We want both balls to be blue, and there are 8 blue balls in the bag. The number of ways to pick 2 blue balls is 8C2 = 28. So n(E) = 28. Using P(E) = n(E)/n(S), P(E) = 28/153 = 28/153.