Solution: (i) Winning of Geeta is a complementary event to winning of Ritu Thus, P(winning of Ritu) + P(winning of Geeta) = 1 P(winning of Geeta) = 1 – P(winning of Ritu) P(winning of Geeta) =...

### (i) If A and B are two complementary events then what is the relation between P(A) and P(B)? (ii) If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?

Solution: (i) Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1. P(A) + P(B) = 1 (ii) P(A) = 0.46 Let P(B) be...

### From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will:(v) be a face card of red colour

(v) There are 26 red cards in a deck, and 6 of these cards are face cards (2 kings, 2 queens and 2 jacks). The number of favourable outcomes for the event of drawing a face card of red color = 6...

### From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will: (i) be a black card. (ii) not be a red card. (iii) be a red card. (iv) be a face card.

(iii) Number of red cards in a deck = 26 The number of favourable outcomes for the event of drawing a red card = 26 Then, probability of drawing a red card = 26/52 = ½ (iv) There are 52 cards...

### From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will: (i) be a black card. (ii) not be a red card.

Solution: We know that, Total number of cards = 52 So, the total number of outcomes = 52 There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are...

### In a single throw of a die, find the probability that the number: (iii) will be an odd number.

(iii) If E = event of getting an odd number = {1, 3, 5} So, n(E) = 3 Then, probability of a getting an odd number = n(E)/ n(s) = 3/6 = ½

### In a single throw of a die, find the probability that the number: (i) will be an even number. (ii) will not be an even number.

Solution: Here, the sample space = {1, 2, 3, 4, 5, 6} n(s) = 6 (i) If E = event of getting an even number = {2, 4, 6} n(E) = 3 Then, probability of a getting an even number = n(E)/ n(s) = 3/6 = ½...

### In a single throw of a die, find the probability of getting a number: (iii) not greater than 4.

(iii) E = event of getting a number not greater than 4 = {1, 2, 3, 4} So, n (E) = 4 Then, probability of getting a number not greater than 4 = n(E)/ n(s) = 4/6 = 2/3

### In a single throw of a die, find the probability of getting a number: (i) greater than 4. (ii) less than or equal to 4.

Solution: Here, the sample space = {1, 2, 3, 4, 5, 6} So, n (s) = 6 (i) If E = event of getting a number greater than 4 = {5, 6} So, n (E) = 2 Then, probability of getting a number greater than 4 =...

### . A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:(v) not a black ball

(v) There are 3 + 2 = 5 balls which are not black So, the number of favourable outcomes = 5 Thus, P(getting a white ball) = 5/10 = ½

### A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is: (iii) a white ball. (iv) not a red ball.

(iii) There are 3 white balls So, the number of favourable outcomes = 3 Thus, P(getting a white ball) = 3/10 = 3/10 (iv) There are 3 + 5 = 8 balls which are not red So, the number of favourable...

### A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is: (i) a black ball. (ii) a red ball

Solution: Total number of balls = 3 + 5 + 2 = 10 So, the total number of possible outcomes = 10 (i) There are 5 black balls So, the number of favourable outcomes = 5 Thus, P(getting a black ball) = ...

### 1. A coin is tossed once. Find the probability of: (i) getting a tail (ii) not getting a tail

Solution: Here, the sample space = {H, T} i.e. n(S) = 2 (i) If A = Event of getting a tail = {T} Then, n(A) = 1 Hence, the probability of getting a tail = n(A)/ n(S) = 1/2 (ii) Not getting a tail As...