10 questions · timed · auto-graded
Solution:
The probability of an event always lies between 0 and 1.
Since $\frac{5}{3}>1,$ it cannot be the probability of an event.
Solution:
If two coins are tossed simultaneously, then possible cases are HH, TH, HT, TT.
Total number of cases = 4
Number of favorable cases (atmost one head) = (HT, TH, TT) = 3
Now,
Probability of getting atmost one head $=\frac{3}{4}$
Solution:
The probability of an impossible event is always zero since the chances of occurring of that event is zero.
Solution:
Probability of an event $=\frac{\text{No. of favorable cases}}{\text{Total no. of cases}}$
Since number of favoravle cases can not be greater than total number of cases,
Probability < 1.
Solution:
The chance of occuring of an certain event is always 100%.
Probability $=\frac{\text{No. of favorable cases}}{\text{Total no. of cases}}$
Thus, for a certain event,
Number of favorable cases = Total number of cases
⇒ Probability = 1
| Class | X | IX | VIII | VII | VI | V |
| Attendance | 30 | 62 | 85 | 92 | 76 | 55 |
Solution:
Total number of classes = 6
Number of classes having attendance > 75% = VIII, VII, VI = 3
⇒ Required probability $=\frac{3}{6}=\frac{1}{2}$
Solution:
Probability that Ronaldo makes a goal
$=\frac{\text{Number of goal made in all kicks}}{\text{Total number of kicks}}$
$=\frac{4}{10}=\frac{2}{5}$
| Outcome | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 200 | 30 | 120 | 100 | 50 | 100 |
Solution:
Prime numbers in 1, 2, 3, 4, 5, 6 are: 2, 3, 5.
Number of times 2, 3, 5 occur = 30 + 120 + 50 = 200
Total number of cases = 200 + 30 + 120 + 100 + 50 + 100 = 600
Required probability $=\frac{\text{Cases when we obtained (2, 3, 5)}}{\text{Total no. of cases}}$
$=\frac{200}{600}=\frac{1}{3}$
Solution:
Probability of getting a tail $=\frac{3}{8}$
⇒ Probability of geeting a head $=1-\frac{3}{8}=\frac{5}{8}$
Also,
Probability of getting a head $=\frac{\text{No.of heads obtained}}{\text{Total no. of trials}}$
$\Rightarrow\ \frac{5}{8}=\frac{\text{No. of heads obtained}}{1000}$
⇒ No. of heads obtainted $=\frac{5}{8}\times1000=625$
Solution:
Prime numbers from 51 to 100:
53, 59, 61, 67, 71, 73, 79, 83, 89, 97
⇒ Number of prime numbers = 10
⇒ Numver of non-prime numbers = 50 - 10 = 40
Total numbers = 50
Thus, probability of getting no-prime number $=\frac{40}{50}=\frac{4}{5}$