Exercise 25B

### If two coins are tossed once, what is the probability of getting: (iii) both heads or both tails.

(iii) E = event of getting both heads or both tails = {HH, TT} n(E) = 2 Hence, probability of getting both heads or both tails = n(E)/ n(S) = 2/4 = ½

### If two coins are tossed once, what is the probability of getting: (i) both heads. (ii) at least one head.

Solution: We know that, when two coins are tossed together possible number of outcomes = {HH, TH, HT, TT} So, n(S) = 4 (i) E = event of getting both heads = {HH} n(E) = 1 Hence, probability of...

### A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.

Solution: In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6} So, n(S) = 6 For two dice, n(S) = 6 x 6 = 36 Favorable cases where the sum is 10 or more with 5 on 1st die = {(5, 5), (5,...

### A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?

Solution: We know that, Number of pages in the book = 85 Number of possible outcomes = n(S) = 85 Out of 85 pages, pages that sum up to 8 = {8, 17, 26, 35, 44, 53, 62, 71, 80} So, pages that sum up...

### A die is thrown once. Find the probability of getting a number: (iii) less than 8 (iv) greater than 6

(iii) On a dice, numbers less than 8 = {1, 2, 3, 4, 5, 6} So, n(E) = 6 Hence, probability of getting a number less than 8 = n(E)/ n(S) = 6/6 = 1 (iv) On a dice, numbers greater than 6 = 0 So, n(E) =...

### A die is thrown once. Find the probability of getting a number: (i) less than 3 (ii) greater than or equal to 4

Solution: We know that, In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6} So, n(S) = 6 (i) On a dice, numbers less than 3 = {1, 2} So, n(E) = 2 Hence, probability of getting a number...

### From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: (iii) 3 and 5 (iv) 3 or 5

(iii) From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e. {15} So, favorable number of events = n(E) = 1 Hence, probability of selecting a card with a multiple of 3 and...

### From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: (i) 3 (ii) 5

Solution: We know that, there are 25 cards from which one card is drawn. So, the total number of elementary events = n(S) = 25 (i) From numbers 1 to 25, there are 8 numbers which are multiple of 3...

### Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is: (v) less than 48

(v) From numbers 1 to 100, there are 47 numbers which are less than 48 i.e. {1, 2, ……….., 46, 47} So, favorable number of events = n(E) = 47 Hence, probability of selecting a card with a number less...

### Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is: (iii) between 40 and 60 (iv) greater than 85

(iii) From numbers 1 to 100, there are 19 numbers which are between 40 and 60 i.e. {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59} So, favorable number of events = n(E)...

### Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is: (i) a multiple of 5 (ii) a multiple of 6

Solution: We kwon that, there are 100 cards from which one card is drawn. Total number of elementary events = n(S) = 100 (i) From numbers 1 to 100, there are 20 numbers which are multiple of 5...

### Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:(iii) an even number and a multiple of 3 (iv) an even number or a multiple of 3

(iii) From numbers 2 to 10, there is one number which is an even number as well as multiple of 3 i.e. 6 So, favorable number of events = n(E) = 1 Hence, probability of selecting a card with a number...