Answer: C Here we can have four cases:- a). x is even, y is even b). x is odd, y is even c). x is even, y is odd d). x is odd, y is odd Out of these four cases , in case 'b' and 'c' the sum will be odd. So required probability = 2/4 = 1/2.
Q. No. 20:
A 5-digit number is formed by the digits 1,2,3,4, and 5 without repetition. What is the probability that the number formed is a multiple of 4?
Answer: D Total number of 5 digits number = 5! = 120. Now to be multiply of 4, the last 2 digits of the number has to be divisible by 4. i.e they must be 12, 24, 32 or 52. Corresponding to each of these ways there are 3! = 6, i.e 6 ways of filling the remaining 3 places. The required probability = (4*6)/120 = 1/5
Q. No. 21:
If five dice are thrown simultaneously, what is the probability of getting the sum as seven?
Answer: A The following are two cases when the sum will be 7 7= 1+1+1+1+3 =5C1 ways =5 7=1+1+1+2+2= 5C2 ways=10 Total number of possible ways of throwing five dice = 65 The required probability = 15/65
Q. No. 22:
Dishant speaks truth in 3/4 cases and Abhishek lies in 1/5 cases. What is the percentage of cases in which both Dishant and Abhishek contradict each other in stating a fact?
Answer: B D and A will contradict each other when one speaks truth and other speaks lies. Probability of D speak truth and A lies = 3/4 * 1/5 = 3/20 Probability of A speak truth and D lies = 4/5 * 1/5 = 1/5 The two prob are mutually exclusive. Hence, probabilities that D and A contradict each other => 3/20 + 1/5 = 7/20 = 35%
Q. No. 23:
X1, x2, x3... x50 are are fifty real numbers such that xr < xr=1 for r = 1, 2, 3... 49. Five numbers out ofthese are picked up at random. The probability that the five numbers have x20 as the middle is:
Answer: B There is a typographical error in the problem statement. Please read xr< xr=1 as xr < xr+1.This means x1< x2 < x3 ......< x50. Hence each of the 19 numbers are less than the number x20.And each of the 30 numbers x21, x22........x50 is greater than x20.Out of the five numbers that are randomly picked from the set , when two numbers are picked from the set {x1, x2.....x19} and two others picked from the set {x21, x22........x50} then the number x20 will always in the middle. When five number are arranged in an order. Total number of ways of selecting such five numbers is (19C2*30C2). As total number of ways of selecting a set of any five number out of 50 is 50C5, Thus the required probability is (19C2*30C2)/50C5
Q. No. 24:
A dice is rolled three times and the sum of the numbers appearing on the uppermost face is 15. The chance that the first roll was a four is :
Answer: B The sum of numbers can be 15 in the following three ways : Case I : 15 = 3+6+6 The first, second and third throws can be (3,6,6), (6,3,6) and (6,6,3) respectively. Total number of ways in which 3,6 and 6 can be obtained = 6 Case II : 15 = 4+5+6 The first, second and third throws can be either 4, 5 and 6. Total number of ways in which 4,5 and 6 can be obtained = 6 Case III : 15 = 5+5+5 The first, second and third throws can be 5,5 and 5. Total number of ways in which 5,5, and 5 can be obtained = 1. Hence, The total number of ways = 3+6+1= 10 The total number of ways in which the first roll will be 4 is 2. Required chance = 2/10 = 1/5.