Probability: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 2)

The bag has 9 white balls and 9 black balls, so 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 the same colour, either both white or both black. The number of ways to pick 2 white balls is 9C2 = 36. The number of ways to pick 2 black balls is 9C2 = 36. Adding these gives n(E) = 36 + 36 = 72. Using P(E) = n(E)/n(S), P(E) = 72/153 = 8/17.

The bag has 10 white balls and 9 black balls, so 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 the same colour, either both white or both black. The number of ways to pick 2 white balls is 10C2 = 45. The number of ways to pick 2 black balls is 9C2 = 36. Adding these gives n(E) = 45 + 36 = 81. Using P(E) = n(E)/n(S), P(E) = 81/171 = 9/19.

The bag has 7 white balls and 4 black balls, so 11 balls in total. We draw 2 balls at random, so n(S) = 11C2 = 55, the number of ways to choose any 2 balls. We want both balls to be the same colour, either both white or both black. The number of ways to pick 2 white balls is 7C2 = 21. The number of ways to pick 2 black balls is 4C2 = 6. Adding these gives n(E) = 21 + 6 = 27. Using P(E) = n(E)/n(S), P(E) = 27/55 = 27/55.

The bag has 5 white balls and 5 black balls, so 10 balls in total. We draw 2 balls at random, so n(S) = 10C2 = 45, the number of ways to choose any 2 balls. We want both balls to be the same colour, either both white or both black. The number of ways to pick 2 white balls is 5C2 = 10. The number of ways to pick 2 black balls is 5C2 = 10. Adding these gives n(E) = 10 + 10 = 20. Using P(E) = n(E)/n(S), P(E) = 20/45 = 4/9.

The bag has 10 white balls and 8 black balls, so 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 the same colour, either both white or both black. The number of ways to pick 2 white balls is 10C2 = 45. The number of ways to pick 2 black balls is 8C2 = 28. Adding these gives n(E) = 45 + 28 = 73. Using P(E) = n(E)/n(S), P(E) = 73/153 = 73/153.

The bag has 5 white balls and 10 black balls, so 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 the same colour, either both white or both black. The number of ways to pick 2 white balls is 5C2 = 10. The number of ways to pick 2 black balls is 10C2 = 45. Adding these gives n(E) = 10 + 45 = 55. Using P(E) = n(E)/n(S), P(E) = 55/105 = 11/21.

The bag has 9 white balls and 4 black balls, so 13 balls in total. We draw 2 balls at random, so n(S) = 13C2 = 78, the number of ways to choose any 2 balls. We want both balls to be the same colour, either both white or both black. The number of ways to pick 2 white balls is 9C2 = 36. The number of ways to pick 2 black balls is 4C2 = 6. Adding these gives n(E) = 36 + 6 = 42. Using P(E) = n(E)/n(S), P(E) = 42/78 = 7/13.

The bag has 9 white balls and 8 black balls, so 17 balls in total. We draw 2 balls at random, so n(S) = 17C2 = 136, the number of ways to choose any 2 balls. We want both balls to be the same colour, either both white or both black. The number of ways to pick 2 white balls is 9C2 = 36. The number of ways to pick 2 black balls is 8C2 = 28. Adding these gives n(E) = 36 + 28 = 64. Using P(E) = n(E)/n(S), P(E) = 64/136 = 8/17.

The bag has 8 white balls and 10 black balls, so 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 the same colour, either both white or both black. The number of ways to pick 2 white balls is 8C2 = 28. The number of ways to pick 2 black balls is 10C2 = 45. Adding these gives n(E) = 28 + 45 = 73. Using P(E) = n(E)/n(S), P(E) = 73/153 = 73/153.

The bag has 6 white balls and 9 black balls, so 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 the same colour, either both white or both black. The number of ways to pick 2 white balls is 6C2 = 15. The number of ways to pick 2 black balls is 9C2 = 36. Adding these gives n(E) = 15 + 36 = 51. Using P(E) = n(E)/n(S), P(E) = 51/105 = 17/35.

A well-shuffled pack has 52 cards, so n(S) = 52. The pack has 4 suits of 13 cards each, with 26 red cards and 26 black cards. Among these, 2 cards are a black queen. So n(E) = 2. Using P(E) = n(E)/n(S), P(E) = 2/52 = 1/26.

A well-shuffled pack has 52 cards, so n(S) = 52. The pack has 4 suits of 13 cards each, with 26 red cards and 26 black cards. Among these, 26 cards are a black card. So n(E) = 26. Using P(E) = n(E)/n(S), P(E) = 26/52 = 1/2.

A well-shuffled pack has 52 cards, so n(S) = 52. The pack has 4 suits of 13 cards each, with 26 red cards and 26 black cards. Among these, 4 cards are a queen. So n(E) = 4. Using P(E) = n(E)/n(S), P(E) = 4/52 = 1/13.

A well-shuffled pack has 52 cards, so n(S) = 52. The pack has 4 suits of 13 cards each, with 26 red cards and 26 black cards. Among these, 13 cards are a diamond. So n(E) = 13. Using P(E) = n(E)/n(S), P(E) = 13/52 = 1/4.

A well-shuffled pack has 52 cards, so n(S) = 52. The pack has 4 suits of 13 cards each, with 26 red cards and 26 black cards. Among these, 4 cards are a king. So n(E) = 4. Using P(E) = n(E)/n(S), P(E) = 4/52 = 1/13.

A well-shuffled pack has 52 cards, so n(S) = 52. The pack has 4 suits of 13 cards each, with 26 red cards and 26 black cards. Among these, 13 cards are a heart. So n(E) = 13. Using P(E) = n(E)/n(S), P(E) = 13/52 = 1/4.

A well-shuffled pack has 52 cards, so n(S) = 52. The pack has 4 suits of 13 cards each, with 26 red cards and 26 black cards. Among these, 26 cards are a red card. So n(E) = 26. Using P(E) = n(E)/n(S), P(E) = 26/52 = 1/2.

A well-shuffled pack has 52 cards, so n(S) = 52. The pack has 4 suits of 13 cards each, with 26 red cards and 26 black cards. Among these, 13 cards are a spade. So n(E) = 13. Using P(E) = n(E)/n(S), P(E) = 13/52 = 1/4.

A well-shuffled pack has 52 cards, so n(S) = 52. The pack has 4 suits of 13 cards each, with 26 red cards and 26 black cards. Among these, 2 cards are a red king. So n(E) = 2. Using P(E) = n(E)/n(S), P(E) = 2/52 = 1/26.

A well-shuffled pack has 52 cards, so n(S) = 52. The pack has 4 suits of 13 cards each, with 26 red cards and 26 black cards. Among these, 13 cards are a club. So n(E) = 13. Using P(E) = n(E)/n(S), P(E) = 13/52 = 1/4.

Each die has 6 faces, and the two dice are thrown together. So the sample space has n(S) = 6 x 6 = 36 equally likely outcomes. We want the outcomes where the dice show a total divisible by 4 or 6. Checking each of the 36 outcomes against this condition gives n(E) = 14 favourable outcomes. Using P(E) = n(E)/n(S), P(E) = 14/36 = 7/18.

Each die has 6 faces, and the two dice are thrown together. So the sample space has n(S) = 6 x 6 = 36 equally likely outcomes. We want the outcomes where the dice show a total of 8. Checking each of the 36 outcomes against this condition gives n(E) = 5 favourable outcomes. Using P(E) = n(E)/n(S), P(E) = 5/36 = 5/36.

Each die has 6 faces, and the two dice are thrown together. So the sample space has n(S) = 6 x 6 = 36 equally likely outcomes. We want the outcomes where the dice show a total of 7. Checking each of the 36 outcomes against this condition gives n(E) = 6 favourable outcomes. Using P(E) = n(E)/n(S), P(E) = 6/36 = 1/6.

Each die has 6 faces, and the two dice are thrown together. So the sample space has n(S) = 6 x 6 = 36 equally likely outcomes. We want the outcomes where the dice show a total more than 7. Checking each of the 36 outcomes against this condition gives n(E) = 15 favourable outcomes. Using P(E) = n(E)/n(S), P(E) = 15/36 = 5/12.

Each die has 6 faces, and the two dice are thrown together. So the sample space has n(S) = 6 x 6 = 36 equally likely outcomes. We want the outcomes where the dice show a total of at least 10. Checking each of the 36 outcomes against this condition gives n(E) = 6 favourable outcomes. Using P(E) = n(E)/n(S), P(E) = 6/36 = 1/6.