NCERT Solution for Probability

(i) Probability of an event E + Probability of the event ‘not E’ + ............ .

(ii) The probability of an event that cannot happen is ............. . Such an event is called ........... .

(iii) The probability of an event that is certain to happen is ................ . Such an event is called ................

(iv) The sum of the probabilities of all the elementary events of an experiment is................ .

(v) The probability of an event is greater than or equal to................ and less than or equal to ................ .

(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event.

(iii) The probability of an event that is certain to happen is 1. Such an event is called sure or certain event.

(iv) The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

(iv) A baby is born. It is a boy or a girl.

(ii) The player may shoot or miss the shot.

∴ The outcomes are not equally likely.

(iii) In advance it is known that the answer is to be either right or wrong.

∴ The outcomes right or wrong are equally likely to occur.

(iv) In advance it is known the newly bom baby has to be either a boy or a girl.

∴ The outcomes either a boy or a girl are equally likely to occur.

(A)

(B) –1.5

(C) 15%

(D) 0.7

∴ (B) –1.5 cannot be the probability of an event.

∴ 0.05 + P_{(not E)} = 1 ⇒ P_{(not E)} = 1 – 0.05

= 0.95

Thus, probability of 'not E' = 0.95.

(i) an orange flavoured candy?

(ii) a lemon flavoured candy?

∴ Taking out any orange flavoured candy is not possible.

⇒ Probability of taking out an orange flavoured candy = 0.

(ii) Also, probability of taking out a lemon flavoured candy = 1.

And the probability of 2 students not having the same birthday = P_{(nSB)}

∴ P_{(nSB)} + P_{(nSB)} = 1

⇒ P_{(SB)} + 0.992 = 1

⇒ P_{(SB)} = 1 – 0.992 = 0.008

So, the required probability of 2 boys having the same birthday = 0.008.

∴ Number of all possible outcomes = 8

(i) For red balls:

∵There are 3 red balls.

∴ Number of favourable outcomes = 3

(iii) not green?

(i) For red marbles:

∵Number of red marbles = 5

∴ Number of favourable outcomes = 5

∴ Probability of red marbles, P(red)

(ii) For white balls:

∵Number of white balls = 8

∴ Probability of white balls,

(iii) For not green balls:

∵Number of white balls = 4

∴ Number of 'not green' balls = 17 – 4 = 13

i.e., Favourable outcomes = 13

∴ Probability of ball 'not green'

50 p coins = 100

Re 1 coins = 50

Rs 2 coins = 20

Rs 5 coins = 10

Total number of coins = 100 + 50 + 20 + 5 = 180

(i) For a 50 p coin:

Favourable events = 100

(ii) For not a Rs. 5 coin:

∵ Number of Rs. 5 coins = 10

∴ Number of 'not Rs. 5' coins = 180 – 10 = 170

⇒ Favourable outcomes = 170

Male fishes = 5

Female fishes = 8

∴ Total number of fishes = 5 + 8 = 13

⇒ Total number of outcomes = 13

Number of favourable outcomes = 5

(i) 8? (ii) an odd number?

(iii) a number greater than 2?

(iv) a number less than 9?

(i) When pointer points at 8:

Total number of outcomes = 8

Number of favourable outcomes = 1

(i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

∴ Number of total outcomes = 6

(i) For prime numbers:

Since 2, 3, and 5 are prime number,

∴ Favourable outcomes = 3

(iii) For an odd number:

Since 1, 3 and 5 are odd numbers.

⇒ Favourable outcomes = 3

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

∴ Total number of possible outcomes = 52

(i) For a king of red colour:

∵Number of red colour kings = 2 [∵ Kings of diamond and heart are red]

∴ Number of favourable outcomes = 2

(ii) For a face card:

∵4 kings, 4 queens and 4 jacks are face cards

∴ Number of face cards = 12

⇒ Number of favourable outcomes = 12

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

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? and (b) a queen?

Number or good pens = 132

Number of detective pens = 12

∴ Total number of pens = 132 + 12 = 144

For good pens:

∵ There are 132 good pens

∴ Number of favourable outcomes = 132

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

∴ Total number of outcomes = 20

(i) ∵ Number defective bulbs = 4

i.e., Favourable outcomes = 4

(ii) ∵ The bulb drawn above is not included in the lot.

∴ Remaining number of bulbs = 20 – 1 = 19.

⇒ Total number of possible outcomes = 19.

∵ Number of bulbs which are not defective = 19 – 4 = 15

⇒ Favourable number of outcomes = 15

Total number of discs = 90

∴ Total number of possible outcomes = 90

(i) For a two-digit number:

Since the two-digit numbers are 10, 11, 12,...., 90.

∴ Number of two-digit numbers = 90 -. 9 = 81

[ ∵ 1, 2, 3, 4, 5, 6, 7, 8, and 9 are 1-digit numbers]

⇒ Number of favourable outcomes = 81

(ii) For a perfect square:

Perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81

∴ Number of perfect numbers = 9 Number of favourable outcomes = 9

⇒�nNumber of favourable outcomes = 9

(iii) For a number divisible by 5:

Numbers divisible by 5 [from 1 to 90] are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90

i.e. There are 18 number (1 to 90) which are divisible by 5.

∴ Number of favourable outcomes = 18

The die is thrown once. What is the probability of getting (i) A? and (ii) D?

∴ Total number of letters = 6

⇒ Number of possible outcomes = 6

(1) For the letter A

∵ Two faces are having the letter A.

∴ Number of favourable outcomes = 2

And, the area of the circle = πr^{2}

(i) She will buy it?

(ii) She will not buy it?

⇒ All possible outcomes = 144

(i) Since there are 20 defective pens

∴ Number of good pens 144 – 20 = 124

⇒ Number of favourable outcomes = 124

∴ Probability that she will buy it

(ii) A student argues that th ere are 11 possible sible outcomes 2 , 3, 4, 5, 6, 7, 8 , 9, 10, 11 and 12.

Therefore, each of them has a probability . Do you agree with this argument? justify your answer.

∴ Following are the possible outcomes:

(ii) No. The number of all possible outcomes is 36 and not 11.

∴ The argument is not correct.

∴ All the possible outcomes are:

H H H, H H T, H T T, T T T, T T H, T H T, T T H, H T H

∴ Number of all possible outcomes = 8

Let the event that Hanif will lose the game be denoted by E.

∴ Favourable events are:

H H T, H T H, T H H, T H T, T T H, H T T

⇒ Number of favourable outcomes = 3

(i) 5 will not come up either time? (ii) 5 will come up at least once?

[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]

∴ All possible outcomes are:

∴ All possible outcomes = 36

(i) Let E be the event that 5 does not come up either time, then

The favourable outcomes are [36 – (5 + 6)] = 25

(ii) Let N be the event that 5 will come up at least once, then Number of favourable outcomes = 5 + 6 = 11

(i) If two coins are tossed simultaneously there are three possible outcomes-two heads, two tails or one of each. Therefore, fir each of these outcomes, the probability is

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

Because, the situation 'one of each' can result in two ways HT and M.

∴ The probability = 4

(ii) Correct.

Because the two outcomes are possible.

= 5 × 5 = 25

(i) For both customers visiting same day:

Number of favourable outcomes = 5

[∵ (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)]

What is the probability that the total score is

(i) even? (ii) 6? (iii) at least 6?

∴ Number of all possible outcomes = 36

(i) For total score being even:

Favourable outcomes = 18

[∵ The even outcomes are: 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8]

(ii) For the score being 6:

In list of score, we have four 6' s.

∴ Favourable outcomes = 4

(iii) For the score being at least 6:

The favourable scores are:

7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12

∴ Number of favourable outcomes = 15

∴ Total number of balls = x + 5

Number of possible outcomes = (x + 5).

For a blue ball favourable outcomes = x

∴ Probability of drawing a blue ball

∴ Number of possible outcomes = 12

Case-I: For drawing a black ball

Number of favourable outcomes = x

∴ Probability of getting a black ball

Now, the total number of balls

= 12 + 6

= 18

⇒ Number of possible outcomes = 18

Since, the number of black balls now

= (x + 6).

∴ Number of possible outcomes = 24.

Let there are x blue marbles in the jar.

∴ Number of green marbles = 24 – x

⇒ Favourable outcomes = (24 – x)

∴ Required probability for drawing a green marble

Now, according to the condition, we have:

Thus, the required number of blue balls is 8.