Six different balls are put in three different boxes, no box being empty. The probability of putting balls in the boxes in equal numbers is,

Q: Six different balls are put in three different boxes, no box being empty. The probability of putting balls in the boxes in equal numbers is,

(A) 3/10

(B) 1/6

(C) 1/5

(D) none of these

Sol. Total number of ways to distribute the balls so that no box is empty are [(1, 1, 4), (2, 1, 3), (2, 2, 2)]

$\large \frac{3!}{2!} [( 6_{C_1} . 5_{C_1} . 4_{C_4} ] + 3! ( 6_{C_2} . 4_{C_1} . 3_{C_3}) + ( 6_{C_2} . 4_{C_2} . 2_{C_2})$

= 90 + 6.60 + 90 = 540

Required probability $\large = \frac{90}{540} = \frac{1}{6}$

A die is thrown three times and the sum of three numbers obtained is 15. The probability of first throw being 4 is

Q: A die is thrown three times and the sum of three numbers obtained is 15. The probability of first throw being 4 is

(A) 1/18

(B) 1/5

(C) 4/5

(D) 17/18

Sol. If first throw is four, then sum of numbers appearing on last two throws must be equal to eleven. That means last two throws are (6, 5) or (5, 6).

Now there are 10 ways to get the sum as 15. [(5, 5, 5) (4, 5, 6) (3, 6, 6)]

⇒ Required probability $\large = \frac{2}{10} = \frac{1}{5}$

A and B play a game of tennis. The situation of the game is as follows; if one scores two consecutive points after a deuce he wins;

Q: A and B play a game of tennis. The situation of the game is as follows; if one scores two consecutive points after a deuce he wins; if loss of a point is followed by win of a point, it is deuce. The chance of a server to win a point is 2/3. The game is at deuce and A is serving. Probability that A will win the match is, (serves are changed after each game)

(A) 3/5

(B 2/5

(C) 1/2

(d) 4/5

Sol. Let us assume that ‘A’ wins after n deuces, n ∈ [(0, ∞)

Probability of a deuce  $\large = \frac{2}{3} . \frac{2}{3} + \frac{1}{3}.\frac{1}{3} = \frac{5}{9}$

= (A wins his serve then B wins his serve or A loses his serve then B also loses his serve).

Now probability of ‘A’ winning the game $\large = \Sigma_{n=0}^{\infty}(5/9)^n (\frac{2}{3})\frac{1}{3}$

$\large = \frac{1}{1-(5/9)}.\frac{2}{9} = \frac{1}{2}$

A bag contains 4 tickets numbered 1, 2, 3, 4 and another bag contains 6 tickets numbered 2, 4, 6, 7, 8, 9…

Q: A bag contains 4 tickets numbered 1, 2, 3, 4 and another bag contains 6 tickets numbered 2, 4, 6, 7, 8, 9. One bag is chosen and a ticket is drawn. The probability that the ticket bears the number 4 is

(A) 1/48

(B) 1/8

(C) 5/24

(D) None of these

Sol. Required probability $\large = \frac{1}{2} \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{6} = \frac{5}{24}$

A four figure number is formed of the figures 1, 2, 3, 5 with no repetitions. The probability that the number is divisible by 5 is

Q: A four figure number is formed of the figures 1, 2, 3, 5 with no repetitions. The probability that the number is divisible by 5 is

(A) 3/ 4

(B) 1/ 4

(C) 1/8

(D) None of these

Sol. The total number of ways in which four-digit numbers can be formed, is 4! = 24. A number is divisible by 5 if at unit’s place, we have 5. Therefore unit’s place can be filled in one way and the remaining 3 places can be filled with the other digits in 3! ways. Hence the total number of numbers divisible by 5 is 3!=6
The required probability = 6/24 = 1/4 .

One hundred identical coins, each with probability p, showing up heads are tossed once…

Q: One hundred identical coins, each with probability p, showing up heads are tossed once. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is

(A) 1/2

(B) 49/101

(C) 50/101

(D) 51/101

Sol. We have 100C50p50 (1 – p)50 = 100C51p51 (1 – p)49

$\large \frac{1-p}{p} = \frac{100 !}{51! \; 49!} \times \frac{50! \; 50!}{100!}$

$\large \frac{1-p}{p} = \frac{50}{51}$

$\large p = \frac{51}{101}$

India play two matches each with West Indies and Australia .In any match the probabilities of India getting points…

Q: India play two matches each with West Indies and Australia .In any match the probabilities of India getting points 0 , 1 and 2 are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is

(A) 0.8750

(B) 0.0875

(C) 0.0625

(D) 0.0250

Sol. Probability of getting at least seven points = Probability of getting 7 points or 8 points = probability of getting 7 points + probability of getting 8 points.

Seven points in four matches can be obtained in the following four different ways:
(2, 2, 2, 1) ; (2, 2, 1, 2); (2, 1, 2, 2); (1, 2, 2, 2)

Hence probability of getting 7 points = 4. (0.50)3 (0.05) = 0.0250

Eight points in four matches can be obtained only in one way i.e. 2, 2, 2, 2, Hence probability of getting 8 points = (0.50)4 = 0.0625

Thus the required probability = 0.0250 + 0.0625 = 0.0875.