Let A and B be events in a sample space S suchthat P(A) = 0.5, P(B) = 0.4 andP(A B) = 0.6. Observe the following lists
List I | List II | ||
(i) | (1) | 0.4 | |
(ii) | (2) | 0.2 | |
(iii) | (3) | 0.3 | |
(iv) | (4) | 0.1 |
The correct match of List I from List II is
A. (i) (ii) (iii) (iv) | (i) (1) (2) (3) (4) |
B. (i) (ii) (iii) (iv) | (ii) (3) (2) (1) (4) |
C. (i) (ii) (iii) (iv) | (iii) (3) (2) (1) (4) |
D. (i) (ii) (iii) (iv) | (iv) (3) (1) (2) (4) |
If a, b and n are natural numbers, then a2n - 1 + b2n - 1 is divisible by
a + b
a - b
a3 + b3
a2 + b2
A bag contains n white and n black balls. Pairs of balls are drawn at random without replacement successively, until the bag is empty. If the number of ways in which each pair consists of one white and one black ballis 14400, then n is equal to
6
5
4
3
The number of five digit numbers divisible by 5 that can be formed using the numbers 0, 1, 2, 3, 4, 5 without repetition is
240
216
120
96
B.
216
The five digit number is formed only when unit place is either 0 or 5.
Case I. When 0 is in unit place, then rest of numbers can be selected in ways.
= 5! = 120
Case II. When 5 is in unit place, then rest of numbers can be selected in ways
= 5! = 120
And we have to subtract those cases in which 0 is in thousand place.
The number of ways = = 24
Required number of ways = 120 + 120 - 24 = 216
If the coefficients of rth and (r + 1)th terms in the expansion of (3 + 7x)29 are equal, then r is equal to
14
15
18
21