Oops! It appears that you have disabled your Javascript. In order for you to see this page as it is meant to appear, we ask that you please re-enable your Javascript!

**Q.1 Â Â Â Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?**

* Sol. Â Â Â Â Â *Elementary events associated to the random experiment of visiting a particular shop in the same week (Tuesday to Saturday) by two customers Shyam and Ekta are â€”

Â Â Â Â Â Â Â Â (T, T), (T, W), (T, TH), (T, F), (T, S)

Â Â Â Â Â Â Â Â (W, T), (W, W), (W, TH), (W, F), (W, S)

Â Â Â Â Â Â Â Â (TH, T), (TH, W), (TH, TH), (TH, F), (TH, S)

Â Â Â Â Â Â Â Â (F, T), (F, W), (F, TH), (F, F), (F, S)

Â Â Â Â Â Â Â Â (S, T), (S, W), (S, TH), (S, F), (S, S)

Â Â Â Â Â Â Â Â where T = Tuesday, W = Wednesday, Th = Thursday, F = Friday and S = Saturday.

Â Â Â Â Â Â Â Â Therefore,Â Total number of elementary events = 5 5 = 25

Â Â Â Â Â Â Â Â Then, elementary events favourable to A are (T, T), (W, W), (TH, TH), (F, F) and (S, S)

Â Â Â Â Â Â Â Â Therefore,Â Favourable number of elementary events = 5

Â Â Â Â Â Â Â Â Hence, required probability =

Â Â Â Â Â Â Â Â Therefore,Â Favourable number of elementary events = 8

Â Â Â Â Â Â Â Â Hence, the required probability =

Â Â Â Â Â Â Â Â Therefore, Favourable number of elementary events = 20

Â Â Â Â Â Â Â Â Hence, required probability = .

**Q.2 Â Â A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two time and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:**

**Â Â Â Â Â Â What is the probability that the total score is **

**Â Â Â Â Â Â (i) even? Â Â Â Â Â Â Â Â Â Â Â (ii) 6? Â Â Â Â Â Â Â Â Â Â Â Â (iii) at least 6?
**

Â Â Â Â Â Â Â Number in first throw

Â Â Â Â Â Â Â Clearly total number of elementary events

Â Â Â Â Â Â Â = 6 6 =36

**Â Â Â Â Â Â Â (i)** Let A be the event of getting total score even. Then, elementary events favourable to A are 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8 and 12

Â Â Â Â Â Â Â Therefore,Â Favourable number of elementary events = 18

Â Â Â Â Â Â Â Hence, P(A) =

**Â Â Â Â Â Â Â (ii)** Let A be the event of getting total score 6. Then, elementary events favourable to A are 6, 6, 6 and 6.

Â Â Â Â Â Â Â Therefore,Â Favourable number of elementary events = 4

Â Â Â Â Â Â Â Hence, P(A) =

**Â Â Â Â Â Â Â (iii)** Let A be the event of the total score is at least 6. Then, elementary events favourable to A are 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12 Â Therefore, Favourable number of elementary events = 15

Hence, P(A) = .Â

**Q.3 Â Â Â A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball double that of a red ball, determine the number of blue balls in the bag.
**

Â Â Â Â Â Â Â Â Therefore,Â Total number of balls in the bag = 5 + x

Â Â Â Â Â Â Â Â Now, = Probability of drawing a blue ball

Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â Â = Probability of drawing a red ball

Â Â Â Â Â Â Â = Â

Â Â Â Â Â Â Â Â But it is given that =

Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Hence, there are 10 blue balls in the bag.Â

**Q.4 Â Â Â A box contains 12 balls out of which x are black. If one ball is drawn at random from the box what is the probability that it will be a black ball ?Â If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
**

Â Â Â Â Â Â Â Therefore,Â Total number of elementary events = 12

Â Â Â Â Â Â Â There are x black balls out of which one can be chosen in x ways.

Â Â Â Â Â Â Â Therefore,Â Favourable number of elementary events = x

Â Â Â Â Â Â Â Hence, = P (getting a black ball) =

Â Â Â Â Â Â Â If 6 more black balls are put in the box, then

Â Â Â Â Â Â Â Total number of balls in the box = 12 + 6 = 18

Â Â Â Â Â Â Â Number of black balls in the box = x + 6

Â Â Â Â Â Â Â Therefore,Â = P (getting a white ball) =

Â Â Â Â Â Â Â It is given that

Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â x + 6 = 3x

Â Â Â Â Â Â Â 2x = 6

Â Â Â Â Â Â Â x = 3Â

**Q.5 Â Â Â A jar contains 24 marbles, some are green and other are blue. If a marble is drawn at random from the jar, the probability that it is green is . and the number of blue balls in the jar.
**

Â Â Â Â Â Â Â Â Therefore,Â Total number of elementary events = 24

Â Â Â Â Â Â Â Â Let there be x green falls

Â Â Â Â Â Â Â Â Therefore,Â Favourable number of elementary evens = x

Â Â Â Â Â Â Â Â Therefore,Â P(G) = But, P(G) = Â Â Â Â Â Â Â Â Â Â Â Â Â [Given]

Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â x = = 16

Â Â Â Â Â Â Â Â Number of green marbles = 16

Â Â Â Â Â Â Â Â Number of blue marbles = 24 â€“ 16 = 8

Contact Us