## 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 = \(\frac{2}{6}=\frac{1}{3}\)

**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 **

**Red?****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:**

- A prime number
- A number lying between 2 and 6
- 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 **

**A?****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 **

**A two-digit number,****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