CBSE Solutions For Class 10 Mathematics Chapter 15 Probability

Class 10 Maths Probability

Question 1. In any situation that has only two possible outcomes, each outcome will have Probability. Find whether it is true or false.
Solution: False.

The probability of each outcome will be \(\frac{1}{2}\), only when the two outcomes are equally likely.

Question 2. A Marble is chosen at random from 6 marbles numbered 1 to 6. Find the Probability of getting a marble having number 2 and 6 on it.
solution:

The favourable Case is to get a marble on which both numbers 2 and are written. But there is no such marble.

So, N(E) = 0 and n(S) = 6

∴ Required probability = \(\frac{n(E)}{n(S)}=\frac{0}{6}=0\)

Question 3. A marble is chosen at random from 6 marbles numbered Ito 6. Find the Probability of getting a marble having number 2 or 6 on it.
Solution:

Here N(E) = 2

and n(S) = 6

∴ Required Probability = \(\frac{2}{6}=\frac{1}{3}\)

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Question 4. It is given that in a group of 3 Students, the probability of 2 students not having the Same birthday is 0.992. What is the probability that the 2 students have the Same birthday.
Solution:

The probability of 2 students not having the Same birthday = 0.992

∴ Probability of 2 students having the same birthday

= 1-0.992

= 0.008

Question 5. A bag Contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

1. Red?
2. Not ved?

Solution:

Total balls = 3+5 = -8

Total possible outcomes of drawing a ball at random from the bag = 8

1. Favorable outcomes of drawing a batt at random red ball =3

∴ The probability of drawing a red ball

⇒ \(\frac{\text { Favourable outcomes of drawing a red ball }}{\text { Total possible outcomes }}\)

⇒ \(\frac{3}{8}\)

2. Probability that the ball drawn is not red = 1- probability that the ball drawn

⇒ \(1-\frac{3}{8}=\frac{5}{8}\)

6. A die is thrown once. Find the probability of getting:

1. A prime number
2. A number lying between 2 and 6
3. An odd number.

Solution:

Possible outcomes in one throw of a die = {1,2,3,4,5,6}

Total possible outcomes = 6

1. Prime numbers = {2,3,5}=3

∴ Probability of getting prime number = \(\frac{3}{6}=\frac{1}{2}\)

2. Numbers lying between 2 to 6 = {3, 4, 5} = 3

∴ Probability of this event = \(\frac{3}{6}=\frac{1}{2}\)

3. Odd numbers = {1,3,5}=3

∴ Probability of this event = \(\frac{3}{6}=\frac{1}{2}\)

Question 7. 12 defective pens are accidentally mixed with 132 good to just look at a pen and tell whether or not it is taken out at random from this lot. Determine the probability Out is a good one.
Solution:

No. of good pens = 132

No. of defective pens = 12

Total pens = 132 +12 = 144

Total Favourable outcomes of drawing a pen = 144

Favourable outcomes of drawing a good pen = 132

∴ Probability of drawing a good pen = \(\frac{132}{144}=\frac{11}{12}\)

Question 8. A child has a die whose Six faces show the letters as

given below; The die is thrown once. what is the probability of getting

1. A?
2. D?

Solution:

Total possible outcomes in a throw of die = 6

1. Favourable outcome of getting A = 2

∴ Probability of getting \(A=\frac{2}{6}=\frac{1}{3}\)

2. Favourable outcomes of getting D = 1

∴ Probability of getting D = \(\frac{1}{6}\)

Question 9. 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

1. A two-digit number,
2. A number divisible by 5.

Solution:

We have, n(S) = 90

1. Let A be the event of getting “a two-digit number”.

∴ Favourable cases are 10,11,12,13,14, …….., 90

∴ n(A) = 81

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{81}{90}=\frac{9}{10}\)

2. let B be the event of getting ་་ a number divisible by 5″.

∴ Favourable Cases are 10, 15, 20, 25, 30, …… 90.

Let there be n in numbers.

∴ T_n = 90

∴ 10+ (n-1)5 = 90

⇒ (n-1)5=80

⇒ n-1 = 16

⇒ n = 17

∴ n(B) = 17

P(B) = \(\frac{n(B)}{n(S)}=\frac{17}{90}\)

Question 10. It is known that a box of 600 electric bulbs Contains 12 defective bulbs. One bulb is taken out at random from this box, what is the probability that it is a non-defective bulb?
Solution:

The number of non-defective bulbs in the box = 600-12=588

So, probability of taking out a non-defective bulb = \(\frac{588}{600}=\frac{49}{50}\)

=0.98