Mathematics teachers manual 9 and 10

4. A natural number is chosen at random from the first 30 natural numbers. What is
the probability that the number chosen is either multiple of 3 or a multiple of 4 ?

Solution:

Let M3 and M4 denote the events of getting a multiple of 3 and multiple of 4
respectively.

Then, S = {1, 2 , 3, ........, 30} ? n(S) = (30 1) 1 = 30

M3 = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30} ? n(M3) = 10
M4 = {4, 8, 12, 16, 20, 24, 28} ? n(M4) = 7
n(M3ˆM4) = {12, 24} ? n(M3ˆM4) = 2

Now, P(M3 or M4) = P(M3) P(M4) P(M3ˆM4)

= n(M3) n(M4) n(M3ˆM4)
n(S) n(S) n(S)

= 10 7 2
30 30 30

= 15 = 1
30 2
So, the probability that the number chosen is either multiple of 3 or a multiple of 4 is 21.

5. If a card is drawn at random from a pack of 52 cards and at the same time a marble

is drawn at random from a bag containing 2 red marbles and 3 blue marbles. Find

the probability of getting a blue marble and a king.
Solution:

Let K, B and R denote the events of getting king card, blue and red marbles respectively.

Then, For drawing a card: n(S1) = 52, n(K) = 4 ?P(K) = n(K) = 4 = 1
n(S) 52 13

For drawing a marble: n(S2) = 2 3 = 5, n(B) = 3 ? P(B) = n(B) = 3
Since the events are independent n(S) 5

? P(K and B) = P(K) u P(B) = 1 u 3 = 3
13 5 65
Hence the required probability is 635.

6. A bag contains 5 black, 7 blue and 4 yellow balls. A ball is drawn at random and it
is replaced, then another ball is drawn. Find the probability that:
i) the first is blue and the second is black
ii) both of them are yellow
iii) both of them are of the same colour
iv) the first is black and the second is yellow
v) both of them are not black

Solution:

Let Bk, Be and Y denote the event of getting a black, blue and yellow balls respectively.
Then, n(Bk) = 5, n(Be) = 7 and n(Y) = 4

297 Vedanta Excel in Mathematics Teachers' Manual - 10

?Total number of balls n(S) = 5 7 4 = 16

Since the ball which is draw at first is replaced to draw another ball, it is an

independent event.

Now,

i) P(Be and Bk) = P(Be ˆ Bk) = P1(Be) u P2(Bk)

= n(Be) u n(BK) = 7 u 5 = 35
n(S) n(S) 16 16 256

ii) P(Both are yellow) = P(YY) = P(YˆY)

= P1(Y) u P2(Y) = n(Y) u n(Y)
n(S) n(S)

= 4 u 4 = 1
16 16 16

iii) P(Both are of same colour) = P(Be Be or Bk Bk or YY)

= P(Be Be) P( Bk Bk) P(YY)

= 7 u 7 5 u 5 4 u 4
16 16 16 16 16 16

= 49 25 16 = 90 45 = 45
256 256 128 128

iv) P(Bk and Y) = P(Bk) u P(Y) = 5 u 146 = 5
16 64

v) P(Bk ‰ Bk) = P1(Bk) u P2(Bk)

= [1 P(Bk)] [1 P(Bk)]

= 1 5 1 5 = 12
16 16 256

7. A can solve 90% of the problems given in the exercise of probability and B can solve

70%. What is the probability that at least one of them will solve a problem selected

at random from the exercise ?

Solution: 90 9 70 7
100 10 100 10
Here, p(A) = = and p(B) = =

Since, the event of solving problem is independent.
9 7
So, p(A and B) = p(A ˆ B) = p(A) u p(B) = 10 u 10 = 63
100
Also, the problem may be solved by both the students A and B.

So, the events are non - mutually exclusive.

Now, p(solving problem by at least one of them) = p(A ‰ B) = p(A) p(B) p(A ˆ B)

= 7 7 63 = 63 = 0.97.
10 10 100 100

8. Three children were born in a family. By drawing a tree diagram, find the following

probabilities.

i) at least two daughters ii) all of them are boys iii) at least a boy

Solution:

Let S and D denote the events of having son and daughter respectively.

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Drawing a probability tree diagram:

(S)= 12

1 S P 3 SSSoP(SSS) = 1 u 1 u 1 = 1
D P 2 2 2 8
(S)= 3 (D)=
P 2 1 SSDoP(SSD) = 1 1 1 = 1
2 12 2 u 2 u 2 8
S P
1 D 2 (D)= 1 P (S)= 2 SDSoP(SDS) = 1 u 1 u 1 = 1
2 2 2 8
(S)= 2 2 S 3
D P
P 1 3 (D)= 1 SDDoP(SDD) = 1 1 1 = 1
2 2 2 8
S 12 u u
D
P (D)= 1 P (S)= 2 DSSoP(DSS) = 1 u 1 u 1 = 1
1 2 2 2 8
3
2 1 S P
(S)= 2 D 3 (D)= 1
P DSDo(DSD) = 1 1 1 = 1
S 2 12 2 u 2 u 2 8
D P
2 (D)= 1 P (S)= 2 DDSoP(DDS) = 1 1 1 = 1
2 2 2 8
2 S 3 u u
D P
3 (D)= 1 DDDoP(DDD) = 1 1 1 = 1
2 2 2 8
2 u u

Now, 1 1 1 1 1
8 8 8 8 2
i) P(at least two daughter) = P(SDD) P(DSD) P(DDS) P(DDD) = =
ii)
P(all are boys) = P(SSS) = 1
8

iii) P(at least a boys) = P(SSS) P(SSD) P(SDS) P(SDD) P(DSS) P(DSD) P(DDS)

= 1 1 1 1 1 1 1 = 7
8 8 8 8 8 8 8 8

Extra questions

1. A box contains the lottery tickets numbered from 3 to 32. If a ticket is drawn at random,
[Ans: 51]
what is the probability of the ticket bearing square or cube number?

2. An ace of diamond is lost from a deck of 52 playing cards and a card is drawn at random.
[Ans: 137]
What is the probability of getting black faced card or ace?

3. A dice is rolled and a coin is tossed at the same time, find the probability of occurring
[Ans: 41]
even number on the dice and head on the coin.

4. A card is draw from a well-shuffled pack of 52 cards and at the same time a marble is
drawn at random from a bag containing 2 green and 3 red marbles of same shape and
[Ans: 110]
size. Find the probability of getting a spade and green marble.

5. A bag contains 1 yellow, 1 black and 1 green balls of same shape and size. Two balls are
drawn at randomly one after another without replacement; show the probabilities of all
possible outcomes in a tree-diagram.

6. A coin is tossed thrice successively. Show the probabilities of all the possible outcomes
1
in a tree diagram and find the probability of getting all heads. [Ans: 8 ]

7. Three children were born in the interval of five years. Find the probability of having the
[Ans: 87]
at least one son by drawing a tree-diagram.

299 Vedanta Excel in Mathematics Teachers' Manual - 10