Test: Statistics Part:- 2 - Mathematics MCQ

Number of permutations with ‘2’ in the first position = 19! 
Number of permutations with ‘2’ in the second position = 10 × 18! 
(fill the first space with any of the 10 odd numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18! ways) 
Number of permutations with ‘2’ in 3^rd position = 10 × 9 × 17! 
(fill the first 2 places with 2 of the 10 odd numbers and then the remaining 17 places with remaining 17 numbers)
and so on until ‘2’ is in 11th place. After that it is not possible to satisfy the given condition, since there are only 10 odd numbers available to fill before the ‘2’. So the desired number of permutations which satisfies the given condition is
19! + 10 x 18! + 10 x 9 x 17! + 10 x 9 x 8 x 16!+ .... + 10! x 9!
Now the probability of this happening is given by

Which is clearly not choices (a),(b) or (c)
Thus, Answer is (d) none of these.

Let C denote computes science study and  M denotes maths study.
P(C on monday and C on wednesday) + p(C on monday, c on tuesday and C on wednesday)
= 1 x 0.6 x 0.4 + 1 x 0.4 x 0.4
= 0.24 + 0.16
= 0.40

given 
Convert into standard and normal variates,

 ...(i)
Now since us know that in standard normal distribution, 

It is given that 
P(odd) = 0.9 p(even)
Now since ∑p(x) = 1
∴ p(odd) + p(even) = 1
⇒ 0.9p(even) + p(even) = 1
⇒ p(even) = 1/1.9 = 0.5263
Now, it is given that p (any even face) is same
i.e p(2) = p(4) = p(6)
Now since,

∴ 
= 1/3 (0.5263)
= 0.1754 
It is given that 

From the diagram,
Pdeclared faulty = pq + (1 − p) (1 − q )

p(multiple of 10% |divisor of 10⁹⁹)

Since 10 = 2.5 
10⁹⁹ = 2⁹⁹ . 5⁹⁹ 
Any divisor of 10⁹⁹ is of the form 2^a . 5^b where 0 ≤ a ≤ 99 and 0 ≤ b ≤ 99.
The number of such possibilities is combination of 100 values of a and 100 values of b = 100 × 100 each of which is a divisor of 10⁹⁹. 
So, no. of divisors of 10⁹⁹ = 100 × 100. 
Any number which is a multiple of 10⁹⁶ as well as divisor of 10⁹⁹ is of the form 2^a . 5^b where 96 ≤  a ≤  99 and 96 ≤ b ≤  99. The number of such combinations of 4 values of a and 4 values of b is 4 × 4 combinations, each of which will be a multiple of 10⁹⁶ as well as a divisor of 10⁹⁹. 
∴ p(multiple of 10⁹⁶|divisor of 10⁹⁹) 

Here, P(A) = 1, P(B) = 1/2
Since A, B are independent events,
∴ P(AB) = P(A)P(B)

Let the time taken for first and second modules be represented by x and y and 
total time = t.
and y and total time = t.
∴ t = x + y is a random variable
Now the joint density function 

which is also called as convolution of f₁ and f₂, abbreviated as f₁* f₂.

Here P(H) = P(T) = 1/2
It's a Bernoulli's trials.
∴ Required probability 

Let the marks obtained per question be a random variable X. It’s probability  distribution table is given below: