Shiwani thought of a two-digit number and divided the number by the sum of the digits of the number. He found that the remainder is 3. Dishant also thought of a two-digit number and divided the number by the sum of the digits of the number. He also found that the remainder is 3. Find the probability that the two digit number thought by Shiwani and Dishant is TRUE?
Answer: B Let two digit number that Shiwani thought be 'xy', where x and y are single digit number. Therefore, 10x+y = p(x+y)+3, where p is a natural number p= (10x+y-3)/(x+y). Also y+x>3 Possible values of x and y for which p is a natural number are:- (x=1, y=5), (x=2, y=3), (x=3, y={1,3,5,9}), (x=4, y=7), (x=5, y={1,2,9}), (x=6,y=Null}), (x=7, y={3,5,8}, (x=8,y=Null), (x=9, y=4). There are 14 such two digit numbers that give a remainder of 3 when divided by the sum of the digits. Probability that Dishant thought of the same number as Shiwani = 1/14.
Q. No. 8:
Find the probability that at least 2 defective tomato are drawn, if 4 tomato are drawn from a box containing 10 tomato of which 3 are defective.
Answer: D The number of ways of choosing 4 squares from 64 is 64C4. The possible split ups as far as colour is concerned and the number of ways of selecting the square are listed below:- White =0, Black= 4 , No. of ways =32C0 * 32C4 White =1, Black= 3 , No. of ways
=32C1 *
32C3 White =2, Black= 2 , No. of ways
=32C2 *
32C2 White =3, Black= 1 , No. of ways
=32C3 *
32C1 White =4, Black= 0 , No. of ways
=32C4 *
32C0 The number of ways in which 3 squares are of one colour and fourth is of opposite colour is P(A) = 2*32C1 * 32C3 = 2(32)[32*31*30]/(2*3) The total number of ways of selecting 4 squares is P(B)= 64C4 = (64*63*62*61)/ (2*3*4) The required probability =P(A)/P(B)= 640/1281
Q. No. 10:
An eight-digit telephone number consists of exactly two zeroes. One of the digit is repeated thrice. Remaining three digits are all distinct. If the first three digits(from left to right) are 987, then find the probability of having only one 9, one 8 and one 7 in the telephone number.
Answer: C Case A:- There is only one 9, one 8 and one 7 in the number. Hence there has to be one digit from {1,2,3,4,5,6} repeated thrice. Total number of ways in which such a number can exist = 6C1 * 5! / (3!*2!).= 60 Case B:- One of the three digit {9,8,7} is repeated thrice. Hence there will be one digit from {1,2,3,4,5,6}. Total number of ways in which such a number can exist =>3C1 * 6C1 * 5!/(3!*2!) =540 Total possible telephone numbers = 60+540 = 600. Probability = 60/600 = 1/10.
Q. No. 11:
Dishant was born between October 6th and 10th (6th and 10th excluding). His year of birth is also known. What is the probability of Dishant being born on a Saturday?
Answer: A Since the year of birth is known, the birthday being on Saturday can have a zero probability. Also since between 6th and 10th, there are three days i.e 7th, 8th and 9th. Therefore, probability of birthday falling on Saturday can be 1/3
Q. No. 12:
There are three similar boxes, containing (i). 6 black and 4 white balls (ii). 3 black and 7 white balls (iii). 5 black and 5 white balls, respectively. If you choose one of the three boxes at random and from that particular box picks up a ball at random, and find that to be black, what is the probability that the ball picked up from the second box?
Answer: B Total number of black balls = 6+3+5 =14. Total number of ways of picking black ball =14 Out of those, 3 are from 2nd box required probability = 3/14