In a non-leap year, the probability of having 53 tuesdays or 53 wednesdays is:
-
$\frac{1}{7}$
-
$\frac{2}{7}$
-
$\frac{3}{7}$
-
none os these.
52 questions across 7 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.
M.C.Q (1 Marks)
12 Q→02True False[1 Marks ]
7 Q→03Fill In The Blanks[1 Marks ]
5 Q→041 Marks Question
6 Q→052 Marks Questions
2 Q→063 Marks Question
10 Q→075 Marks Questions
10 Q→One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
In a non-leap year, the probability of having 53 tuesdays or 53 wednesdays is:
$\frac{1}{7}$
$\frac{2}{7}$
$\frac{3}{7}$
none os these.
6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is:
$\frac{1}{432}$
$\frac{12}{431}$
$\frac{1}{132}$
none of these.
The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then $\text{P}(\bar{\text{A}})+\text{P}(\bar{\text{B}})$ is:
A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is:
$\frac{1}{3}$
$\frac{4}{11}$
$\frac{2}{11}$
$\frac{3}{11}$
While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours:
$\frac{29}{52}$
$\frac{1}{2}$
$\frac{26}{51}$
$\frac{27}{51}$
| a. | If E1 and E2 are the two mutually exclusive events | i. | $\text{E}_1\cap\text{E}_2=\text{E}_1$ |
| b. | If E1 and E2 are the mutually exclusive and exhaustive events | ii. | $(\text{E}_1-\text{E}_2)\cup(\text{E}_1\cap\text{E}_2)=\text{E}_1$ |
| c. | If E1 and E2 have common outcomes, then | iii. | $\text{E}_1\cap\text{E}_2=\phi,\text{ E}_1\cup\text{E}_2=\text{S}$ |
| d. | If E1 and E2 are two events such that $\text{E}_1\subset\text{E}_2$ | iv. | $\text{E}_1\cap\text{E}_2=\phi$ |
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