Class 10 Mathematics

Probability

Exercise 15.1 (NCERT) Part 2

Question 14: One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red colour

Solution: Total number of events = 52

Number of red king = 2

P(Red King) `=2/52=1/26`


(ii) A face card

Solution: Number of face cards in a pack = 12

P(Face cards) `=12/52=3/13`

(iii) A red face card

Solution: Number of red face cards in a pack = 6

P(Red face cards) `=6/52=3/26`

(iv) Jack of hearts

Solution: Number of jack of hearts = 1

P(Jack of hearts) `=1/52`

(v) A spade

Solution: Number of spade cards in a pack = 13

P(spade cards) `=13/52=1/4`

(vi) Queen of diamonds

Solution: Number of queen of diamonds in a pack = 1

P(Queen of diamonds) `=1/52`

Question 15: Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

Solution: Total number of events = 5

Number of queen = 1

P(Queen) `=1/5`

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is

(a) an ace?

Solution: Total number of events = 4

P(ace) `=1/4`

(b) a queen?

Solution:Now, there is no queen in the pack

P(Queen) = 0


Question 16: 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Solution: Total number of events = 132 + 12 = 144

Number of favourable events = 132

P(E) `=(132)/(144)=11/12`

Question 17: (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

Solution: Total Number of events = 20

Number of favourable events = 4

P(E) `=(4)/(20)=1/5`

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Solution: Total number of events = 19

Number of favourable events = 15

P(E) `=(15)/(19)`

Question 18: A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number

Solution: Total number of events = 90

Number of favourable events = 90 – 9 = 81

(single digit numbers are from 1 to 9)

P(E) `=(81)/(90)=9/10`

(ii) a perfect square number

Solution: Number of favourable events can be counted by making following list of square numbers:

4, 9, 16, 25, 36, 49, 64, 81

P(E) `=(8)/(90)=4/45`

(iii) a number divisible by 5.

Solution: The largest number divisible by 5 is 90

Number of numbers divisible by 5 can be counted as follows:

`=(90-5)/(5)+1=18`

P(E) `=(18)/(90)=1/5`


Question 19: A child has a die whose six faces show the letters as given below: ABCDEA. The die is thrown once. What is the probability of getting

(i) A?

Solution: Total number of events = 6

Number of A = 2

P(A) `=2/6=1/3`

(ii) D?

Solution: Number of D = 1

P(D) `=1/6`

Question 20: Suppose you drop a die at random on the rectangular region shown below. What is the probability that it will land inside the circle with diameter 1m?

10 probability exercise soluion

Solution: Area of rectangle = 6 sq m

Area of circle

`=πr^2`
`= πxx0.5^2=0.25 π`

Area of rectangle gives the total number of events and area of circle gives the number of favourable events.

P(E) `=(0.25 π)/(6)`

`=(25xx22)/(6xx100xx7)=11/84`

Question 21: A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it?

Solution: Total number of events = 144

Number of good pen = 124

P(buy) `=(124)/(144)=31/36`

(ii) She will not buy it?

Solution: Number of defective pen = 20

P(no buy) `=(20)/(144)=5/36`

Question 22: A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Solution: Possible outcomes of 3 tosses of coin

1st = HHH
2nd = HHT or HTH or THH
3rd = TTH or THT or HTT
4th = TTT

Total number of events = 8

Number of favourable events = 6 (refer to 2nd and 3rd case)

P(E) `=6/8=3/4`

Question 23: A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

Solution: Possible outcomes of 2 throws of a die can be shown by following table:

1, 11, 21, 31, 41, 51, 6
2, 12, 22, 32, 42, 52, 6
3, 13, 23, 33, 43, 53, 6
4, 14, 24, 34, 44, 54, 6
5, 15, 25, 35, 45, 55, 6
6, 16, 26, 36, 46, 56, 6

Total number of events = 36

Number of times when 5 does not come up either of times = 25

P(E) `=25/36`

(ii) 5 will come up at least once?

Solution: Number of times 5 comes at least once = 11

P(E) `=11/36`

Question 24: Which of the following arguments are correct and which are not correct? Give reasons for your answer.

(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3

Solution: Possible outcomes: HH, HT, TH, TT

P(2 heads) `=1/4`

P(2 tails) `=1/4`

P(both) `=2/4=1/2`

So, given answer is incorrect.

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is ½

Solution: This is correct. Number of both outcomes is equal.


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Exercise 15.1 Part 1

Exercise 15.2