Answer: Option C. -> \(\frac{1}{26}\)

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{2}{52}=\frac{1}{26}.\)

Answer: Option C. -> \(\frac{2}{91}\)

Let S be the sample space.

Then, n(S)  = number of ways of drawing 3 balls out of 15

= ¹⁵C₃

_{= \(\frac{(15\times14\times13)}{(3\times2\times1)}\)}

_{= 455.}

Let E = event of getting all the 3 red balls.

So,  n(E) = ⁵C₃ = ⁵C₂ =  \(\frac{(4\times5)}{(2\times1)}\)  = 10.

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{10}{455}=\frac{2}{91}.\)

Answer: Option D. -> \(\frac{13}{102}\)

Let S be the sample space.

Then, n(S) = ⁵²C₂ =   \(\frac{(52\times51)}{(2\times1)}\) = 1326.

Let E = event of getting 1 spade and 1 heart.

So, n(E)  = number of ways of choosing 1 spade out of 13 and 1 heart out of 13

 = (¹³C₁ x ¹³C₁)

= (13 x 13)

= 169.

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{169}{1326}=\frac{13}{102}.\)

Answer: Option B. -> \(\frac{3}{13}\)

Clearly, there are 52 cards, out of which there are 12 face cards.

So, P (getting a face card) =  \(\frac{12}{52}=\frac{3}{13}.\)

Answer: Option B. -> \(\frac{4}{7}\)

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) =  \(\frac{8}{14}=\frac{4}{7}\)

Answer: Option A. -> 8
:
A
Let n be the least number of bombs required and x the number of bombs that hit the bridge. Then x follows a binomial distribution with parameter n and $p=\frac{1}{2}$
Now, $P(X\ge 2)>0.9\Rightarrow 1-P(X<2)>0.9$
$\Rightarrow P(X=0)+P(X=1)<0.1$
$\Rightarrow {}^n{C}_0{\left(\frac{1}{2}\right)}^n+{}^n{C}_1{\left(\frac{1}{2}\right)}^{n-1}\left(\frac{1}{2}\right)<0.1\Rightarrow 10(n+1)<2^n$
This gives $n\ge 8$

Answer: Option C. -> 1013
:
C
Since there are 12 face cards in a deck of 52cards, the probability of drawing a face card is $\frac{12}{52}=\frac{3}{13}$
Hence, the probability of not picking a face card = $1-\frac{3}{13}=\frac{10}{13}$

Answer: Option B. -> 1.0001
:
B
The probability of an event E, P(E)$=\frac{\text{favourable no. of outcomes}}{\text{total no. of outcomes}}$
Since the favourable no. of outcomes are always less than or equal to the total no. of outcomes, $P\left(E\right)\le 1$.

Answer: Option C. -> 16
:
C
Given,
Sample space for rolling a pair of dice = 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) }
$\Rightarrow$ Total number of outcomes = 36
From the sample space, (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1) gives sum of 7.
$\Rightarrow$ Number of favourable outcomes = 6
Probability of an event, P(E) = $\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
P ( getting sum 7) =$\frac{6}{36}=\frac{1}{6}$
Therefore,the probability of getting a sum of 7 on rolling a pair of dice is $\frac{1}{6}$ .

Answer: Option A. -> 0.87
:
A
There are always two possibilities with an event,either it will happen or it will not happen.
We know that the sum of probabilities of all possible outcomes of an event is 1.
$\therefore$ The sum of probabilities of an event happening and an event not happening= 1
Let P(E) bethe probability of an event happening.
(Then, P(not E) is the probability of the event not happening)
Then, P(E) = 0.13(Given)
$\Rightarrow$ P(not E) = 1 – 0.13 = 0.87