Solved Objective Question on Probability Set 1

Question.1

The decimal parts of the logarithms of two numbers taken at random are found to six places. Probability that second can be subtracted from one first without borrowing is,.

(A)

(B)
(C)

(D) none of these

Solution

P₁ =.x₁x₂x₃…. x₆
P₂ =.y₁y₂…..y₆
Any digit after decimal can be 0, 1, 2, - - -9. P₂ can subtract from P₁ without borrowing if x_i ≥ y_i .

Let y_i = l (0 £ l £ 9)
&⇒ x_i = l, l + 1, L, 9

&⇒ After choosing y_i, x_i can be chosen in (10 - l) ways.

&⇒ Required probability =

Question.2

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

Solution

Let us assume that 'A' wins after n deuces, n ∈ [( 0, ¥ )
Probability of a deuce =
(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
=

Question.3

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

(D) 17/18

Solution

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 =

Question.4

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

Solution

Total number of ways to distribute the balls so that no box is empty are [(1, 1, 4), (2, 1, 3), (2, 2, 2)] =
90 + 6.60 + 90 = 540
required probability = .

Question.5

Fifteen coupons are numbered 1, 2, 3, - - - 15. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on the selected coupon is 9, is .

(A)

(B)
(C)

(D) none of these

Solution

Total ways = 15⁷
For favorable ways, we must 7 coupons numbered from 1 to 9 so that '9' is selected atleast once. Thus total number of favorable ways are, 9⁷ – 8⁷

&⇒ Required probability =

Question.6

A number of six digits is written down at random. Probability that sum of digits of the number is even is .

(A)1/2

(B) 3/8

(D) none of these

Solution

As we are considering all possible six digit numbers, half of these would have sum of their digits to be even and half of these would have sum of their digits to be odd. Hence the required probability would be 1/2.

Question.7

Two small squares on a chess board are chosen at random. Probability that they have a common side is,
(A) 1/3 .

(B) 1/9
(C) 1/18

(D) none of these

Solution

There are 64 small squares on a chess board.

&⇒ Total number of ways to choose two squares = ⁶⁴C₂ = 32.63 .

For favourable ways we must chosen two consecutive small squares for any row or any columns

&⇒ Number of favorable ways = (7.8)2

&⇒ Required probability =

Question.8

A bag contains 17 tickets numbered 1 to 17. A ticket is drawn and replaced, then one more ticket is drawn and replaced. Probability that first number drawn is even and second is odd is .

(A)

(B)
(C)

(D) none of these

Solution

Let A be the event that first drawn number is even and B be event that the second drawn number is odd.

Total number of even numbers = = 8

&⇒ P(A) = and P(B) =
&⇒ P(A ÇB) =

Question.9

A student appears for test I, II and III. The student is successful if he passes either in test I, II or I, III. The probability of the student passing in test I, II and III are respectively p. q and ½. If the probability of the student to be successful is ½ then .

(A) p = q = 1

(B) p = q = ½
(C) p = 1, q = 0

(D) p = 1, q = ½

Solution

Let p₁ p₂ and p₃ be the events the student passes test I, II and III respectively,

&⇒ P(p₁) = p, P(p₂) = q, P(p₃) = ½
A be the even of student being successful,

&⇒ P(A) = P(p₁) . P(p₂) + P(p₁). P(p₃)

&⇒

Now out of four given alternatives only p = 1 and q = 0 satisfies this equation. .

Question.10

Consider a set 'P' containing n elements. A subset 'A' of 'P' is drawn and thereafter set 'P' is reconstructed. Now one more subset 'B' of 'P' is drawn. Probability of drawing sets A and B so that A Ç B has exactly one element is
(A) (3/4)ⁿ .n .

(B) n. (3/4)^n-1
(C) n. (3/4)ⁿ .

(D) none of these

Solution

Let x_i be any element of set P, we have following possibilities;
(i) x_i ∈ A, x_i ∈ B
(ii) x_i ∈ A, x_i ∈ B
(iii) x_i ∈ A, x_i ∈ B
(iv) x_i ∈ A, x_i ∈ B

Clearly, the element x_i∈A Ç B if and only if it belongs to A and B both. .

Thus out of these 4 ways only first way is favorable. Now the element that we want to be in the intersection can be chosen in 'n' different ways. Hence required probability is n. (3/4)ⁿ.