Test: Probability - 2 (January 2) - CAT MCQ

Total count of numbers, n(S) = 10

Prime numbers in the given range are 2,3,5 and 7
Hence, total count of prime numbers in the given range, n(E) = 4

Total number of balls = 3 + 5 + 4 = 12

Let S be the sample space.
n(S) = Total number of ways of drawing 3 balls out of 12 = ¹²C₃

Let E = Event of drawing 3 different coloured balls

To get 3 different coloured balls,we need to select one black ball from 3 black balls,
one red ball from 5 red balls, one blue ball from 4 blue balls

Number of ways in which this can be done = ³C₁ × ⁵C₁ × ⁴C₁
i.e., n(E) = ³C₁ × ⁵C₁ × ⁴C₁

Total number of outcomes possible when a coin is tossed = 2 (? Head or Tail)
Hence, total number of outcomes possible when 5 coins are tossed, n(S) = 2⁵

E = Event of getting exactly 2 heads when 5 coins are tossed
n(E) = Number of ways of getting exactly 2 heads when 5 coins are tossed = ⁵C₂

Total number of cards, n(S) = 52

Total number of "Queen" cards, n(E) = 4

Total number of cards = 52

Total number of Ace cards = 4

Total number of outcomes possible when a die is rolled = 6 (? any one face out of the 6 faces)

Hence, total number of outcomes possible when two dice are rolled = 6 × 6 = 36

To get a sum of 7, the following are the favourable cases.
(1, 6), (2, 5), {3, 4}, (4, 3), (5, 2), (6,1)

=> Number of ways in which we get a sum of 7 = 6

To get a sum of 11, the following are the favourable cases. (5, 6), (6, 5) => Number of ways in which we get a sum of 11 = 2

Here, clearly the events are mutually exclusive events. By Addition Theorem of Probability, we have P(a sum of 7 or a sum of 11) = P(a sum of 7) + P( a sum of 11)

Total number of cards = 52

Total Number of King Cards = 4

Total Number of Diamond Cards = 13

Total Number of Cards which are both King and Diamond = 1

Here a card can be both a Diamond card and a King. Hence these are not mutually exclusive events. (Reference : mutually exclusive events) . By Addition Theorem of Probability, we have P(King or a Diamond) = P(King) + P(Diamond) – P(King and Diamond)

The correct answer is B as 
probability of one of them being selected, imply either John must be selected and Dani must not or Dani must be selected and John must not.
The probability of John's selection is 1/3, so that of his disselection is (1-1/3) = 2/3.
The probability of Dani's selection is 1/5, so that of his disselection is (1-1/5) = 4/5.
The probability of John's selection and Dani's disselection is 1/3 x 4/5 = 4/15.
The probability of Dani's selection and John's disselection is 2/3 x 1/5 = 2/15.
So the required probability is 4/15 + 2/15 = 6/15 = 2/5.