Probability Exercise 2

A and B will contradict each other if one speaks truth and other false . So, the required

Probability

4 students out of 5 can be selected in ⁵C₄ ways.

Probability of a student being not a swimmer ⅕

Probability of a student being a swimmer

Required probability

It is given that last 3 digits are randomly dialled. Then each of the digits can be selected out of 10 digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) in 10 ways.

Hence the required probability

Two squares out of 64 can be selected in

The number of ways of selecting those pairs which have a side in common

= ½ (4×2+24×3+36×4) = 112

[Since each of the corner squares has two neighbours each of 24 squares in border rows, other than corner ones has three neighbours and each of the remaining 36 squares have four neighbours and in this computation, each pair of squares has been considered twice].

Hence required probability

=

The number of ways of arranging 50 books

= ⁵⁰P₅₀ = 50!.

The number of ways of choosing places for the five volume dictionary

= ⁵⁰C₅

and the number of ways of arranging the remaining 45 books

= ⁴⁵C₄₅= (45)!

Thus the number of favourable ways

= (⁵⁰C₅) (45 !).

Hence the probability of the required event

Chandra hits the target 4 times in 4 shots. Hence, he hits the target definitely.

The required probability, therefore, is given by.

P(both Atul and Bhola hit) + P(Atul hits, Bhola does not hit) + P(Atul does not hit, Bhola hits)

=

Total number of balls

= 5 + 7 + 8 = 20

Probability that the first ball drawn is white

If balls are drawn with replacement, all the four events will have equal probability.

Therefore, required probability

Let A be the event of getting an odd number.

Here, n (S) = 6 and

n (A) = 3

Probability of getting an odd number

Hence, probability of not getting an odd number

Required probability of 5 successes

Total no. of numbers

= 6 positive + 8 negative = 14

n(S) = ¹⁴C₄

The product of four numbers could be positive when,

(a) all the four numbers chosen are positive or

(b) all the four numbers chosen are negative or

(c) two of the chosen numbers are positive and two are negative.

Required Prob.

=

=

Total no. of outcomes when two dice are thrown = n (S) = 36 and the possible cases for the event that the sum of numbers on two dice is a prime number, are

(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 1), (5, 6), (6, 1), (6, 5)

Number of outcomes favouring the event

= n (A) = 15

Required probability

Probability that the trouser is not black = ⅔

Probability that the shirt is not black = ¾

Since, picking of a shirt and a trouser are independent,

required probability

Exhaustive number of cases = ²⁴C₁₄

Favourable cases = ²²C₁₄

The word ‘society’ contains seven distinct letters and they can be arranged at random in a row in

⁷P₇ ways, i.e. in 7! = 5040 ways.

Let us now consider those arrangements in which all the three vowels come together. So in this case we have to arrange four letters. S,C,T,Y and a pack of three vowels in a row which can be done in

⁵ P₅ i.e. 5! = 120 ways.

Also, the three vowels in their pack can be arranged in

³P₃ i.e. 3! = 6 ways.

Hence, the number of arrangements in which the three vowels come together is

120 × 6 = 720

∴ The probability that the vowels come together

=

Probability of selecting a month = ¹⁄₁₂.

13th day of the month is Friday if its first day is Sunday and the probability of this = ¹⁄₇.

∴ Required probability

=.

A leap-year has 366 days i.e. 52 complete weeks and two days more these two days be two consecutive days of a week. A leap year will have 53 Sundays if out of the two consecutive days of a week selected at random one is a Sunday.

Let S be the sample space and E be the event that out of the two consecutive days of a week one is Sunday, then

S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}

∴ n (S) = 7 and E = {(Sunday, Monday), (Saturday, Sunday)

∴ n (E) = 2

Now, required Probability, P(E) = = ²⁄₇

For each toss there are four choices

(i) A gets head, B gets head

(ii) A gets tail, B gets head

(iii) A gets head, B gets tail

(iv) A gets tail, B gets tail

thus, exhaustive number of ways

= 4⁵⁰.

Out of the four choices listed above (iv) is not favourable to the required event in a toss.

Therefore favourable number of cases is 3⁵⁰.

Hence, the required probability =

If six coins are tossed, then the total no. of outcomes

= (2)⁶ = 64

Now, probability of getting no tail = ¹⁄₆₄

Probability of getting at least one tail

Probability of occurrence of head in a toss of a coin is 1/2.

Required probability = Prob. [Head appears once] + Prob.[Head appears thrice] + Prob. [Head appears five times]

=

= [5 + 10 + 1]

=