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
- 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.
∴ Tn = 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