MATHS M.C.Q (1 Marks)

3. $\frac{1}{9}$

Solution:

When two dice are thrown, there are (6 × 6) = 36 outcomes.

The set of all these outcomes is the sample space given by

S = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

i.e. n(S) = 36

Let E be the event of getting a total score of 5.

Then E = {(1, 4), (2, 3), (3, 2), (4, 1)

$\therefore$ n(E) = 4

Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{4}{36}=\frac{1}{9}$

1. $\frac{\ ^{7}\text{P}_5}{7^5}$​

Solution:

Since, it is an eight - storey building.

So, there are 7 possible options for them in 7 floors in total if ground floor is not considered.

Hence, total possible outcomes = 7× 7× 7 × 7× 7= 7⁵

Thus, number of ways in which 5 persons can leave from seven floors differently $=\ ^{7}\text{P}_5$

Required probability $=\frac{\ ^{7}\text{P}_5}{7^5}$

4. $\frac{23}{24}$

Solution:

Total number of ways of placing four letters in 4 envelops = 4 = 24

All the letters can be dispatched in the right envelops in only one way. Therefore, the probability that all the letters are placed in the right envelops is $\frac{1}{24}$.

Hence, probability that all the letters are not placed in the right envelops $=1-\frac{1}{24}=\frac{23}{24}$

2. Is false.

Solution:

Since the events A, B and C are mutually exclusive, we have:

$\text{P}(\text{A}\cup\text{B}\cup\text{C})=\frac{2}{3}+\frac{1}{4}+\frac{1}{6}=\frac{13}{12}>1$

which is not possible.

Hence, the given statement is false.

1. $\frac{1}{26}$

Solution:

Favourable number of outcomes, with 10 of black suit = 2

Total number of outcomes = 52

Thus, probabilit

$=\frac{2}{52}=\frac{1}{6}$