10th probability illustrative solved examples

ILLUSTRATIVE EXAMPLES

Example 1. An unbiased die is thrown. What is the probability of getting :

(i) an odd number (ii) a multiple of 3 (iii) a perfect square number (iv) a number less than 4.

Solution.

An unbiased die is thrown we may get 1, 2, 3,4,5,6

So,  total number of all possible outcomes= 6

(i) Favourable outcomes for an odd number are 1, 3, 5.

So, no. of favourable outcomes = 3

P (an odd number) = No. of favourable outcomes/ Total no. of possible outcome= 3/6=1/2

(ii) favourable outcomes for a multiple of 3  are 3 and 6.

So, no. of favourable outcomes = 2

P (a multiple of 3) = 2/6= 1/3

(iii) favourable outcomes for ) a perfect square number  are 1 and 4.

So, no. of favourable outcomes = 2

 P (a perfect square number)  =2/6= 1/3

(iv) favourable outcomes for a number less than 4. are 1, 2 and 3.

So, no. of favourable outcomes = 3

P (a number less than 4) = 3/6 =1/2

2. Three unbiased coins are tossed together. Find the probability of getting :

(i) all heads (ii) two heads (iii) one head (iv) at least two heads

Solution. When three unbiased coins are tossed together, possible outcomes are

HHH, HHT, HTH, HTT, THH, THT, TTH and  TTT.       
So, total no. of possible outcomes = 8

(i) favourable outcome = HHH

So, No. of favourable outcome = 1

P (all heads)  =no. of favourable outcomes/Total no. possible outcomes= 1/8

(ii) favourable outcomes are HHT, THH and HTH.

So, no. of favourable outcomes = 3

P (two heads) = 3/8

(iii) favourable outcomes are HTT, THT and TTH.

 So, no. of favourable outcomes = 3       P (one head) =  3/8

(iv) favourable outcomes are HHH, HHT, HTH and THH.

So, no. of favourable outcomes = 4

P (at least two heads) 4/8 =1/2

Example 3. Find the probability that a leap year selected at random will contain 53 Sundays.

Solution. In a leap year, there are 366 days. But  366 days = 52 weeks + 2 days.

Thus, a leap year has always 52 sundays.

The remaining 2 days can be :

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

Clearly, there are seven elementary events associated with this random experiment.

Let E be the event that a leap year has 53 sundays. Clearly, the event E will happer if the last two

days of the leap year are either Sunday and Monday or Saturday and Sunday.

Favourable no. of elementary events = 2

Hence, required probability = 2/7 

Example 4. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn.

Find the probability that the card drawn is :

(i) an ace (ii) either red or king  (iii) a face card (iv) a red face card

Solution. here, total no. of possible outcomes = 52.

(i) There are 4 ace cards in a pack of 52 cards. One ace can be chosen in 4 ways.

So, favourable no. of outcomes = 4

P (an ace) no. = of favourable outcomes/Total no. of possible outcomes= 4/52=1/13

(ii) There are 26 red cards, including 2 red kings. Also, there are 4 kings, two red and two black.

card drawn will be a red card or a king if it is any one of 28 cards (26 red cards and 2 black kings)

So, favourable no. of outcomes = 28

        P(either red or king) =28/52= 7/13

(iii) Kings, queens and jacks are the face cards.

So, favourable no. of outcomes = 3 × 4 = 12

      P(a face card) = 12/52 =3/13

(iv)  There are 6 red face cards, 3 each from diamonds and hearts.

So, favourable no. of outcomes = 6

                      P(a red face card) = 6/52= 3/26

Example 5. Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number of the card is :

(i) an even number                                     (ii) a number less than 14 

(iii) a number which is a perfect square (iv) a prime number less than 20.

Solution.

From 2 to 101, these are (101–2) + 1 = 100 numbers.

So, total no. of possible outcomes = 100.

(i) From 2 to 101, the even numbers are 2, 4, 6, ...., 100 which are 50 in number.

So, number of favourable outcomes = 50

     P(an even number) = no. of favourable outcomes/Total no. of possible outcomes

                                       = 50/100 = 1/2

(ii) From 2 to 101, the numbers less than 14 are 2, 3, ...., 13 which are 12 in number.

So, no. of favourable outcomes    = 12

              P(a number less than 14) = 12/100 = 3/25

(iii) From 2 to 101, the perfect squares are 4, 9, 16, ..... 100, which are 9 in number.

              So, no. of favourable outcomes = 9

 P (a number which is a perfect square) = 9/100

(iv) From 2 to 101, the prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19 which are 8 in

number.

So, no. of favourable  outcomes  = 8

        P (a prime no. less than 20) = 8/100=2/25

Example 6. A bag contains 3 red balls and 5 black balls.A ball is drawn at random from a bag. What is the

probability that the ball drawn is :  (i) red (ii) not red

Solution. Total number of balls = 3 + 5 = 8

(i) P (red ball) = no. of red balls /Total no.of balls=3/8

(ii) P (not red ball) = 1 – P(red ball) =1 – 3/8 =5/8

Example  7. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3 . Find the number of blue marbles in the jar.

Solution. Total number of elementary events = 24.

Let there be x green marbles.

P (green marbles is drawn) = x/24

but, P(green marbles is drawn)= 2/3 (given)

So, x / 24  = 2/3         x=24x2/3                       x   =   16

 Number of green marbles = 16

Number of blue marbles = 24 – 16 = 8 Ans.

Example 8. Suppose you drop a die at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with diameter 1 m?

 Solution. Total area of rectangular region = 3 m × 2 m = 6 m2

Area of the circle = p r²  = p (1/2)² m²        =  p/ 4 m²

P (die to land inside the circle) = p/ 4 m² ÷  6 = p/24 

Example 9. A bag contains 12 balls out of which x are white.

(i) If one ball is drawn at random, what is the probability that it will be a white ball?

(ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be

double than that in (i). Find x.

Solution. (i) Total number elementary events = 12.

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

So, favourable number of elementary events = x

 p1 = P (white ball) = no. of favourable outcomes / Total no. of possible outcomes = x /12

 (ii) If 6 more white balls are put in the bag, then total number of balls in the bag =12 + 6 = 18

and, no. of white balls in the bag = (x + 6)

P₂ = P (getting a white ball) = ( x + 6 )/ 18

It is given that, p₂  =  2 p₁

( x + 6 )/ 18 = 2 ( x / 2)

 x = 3

Example 10. Two dice are thrown simultaneously. Find the probability of getting :

(i) a doublet i.e. same number on both dice. (ii) the sum as a prime number.

Solution. Possible outcomes associated to the random experiment of throwing two dice are :

(1, 1), (1, 2), ......., (1, 6)

(2, 1), (2, 2), ......., (2, 6)

.......................................

.......................................

(6,1) , (6, 2), ......., (6, 6)

Total number of possible outcomes = 6 × 6 = 36

(i) The favourable outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

Total no. of favourable outcomes = 6

So, P(a doublet) = no. of favourable outcomes/Total no. of possible outcomes = 6/36= 1/6

(ii) Here, favourable sum (as a prime number) are 2, 3, 5, 7 and 11.

So, favourable outcomes are (1, 1), (1, 2), (2, 1), (1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2),

(3, 4), (4,3), (6, 5) and (5, 6).

no. of favourable outcomes = 15

P (the sum as a prime number) = 15/36 = 5/12