155E - PROBABILITY

MATH SHORT CUT TRICKS TO SOLVE PROBABILITY

Most of the candidates get confused and choose wrong option of a question because of superficial concept of PROBABILITY question. These are very easy and time saving questions need to be solved within 2-3 seconds by ONE SIMPLE RULE.

ONE SIMPLE RULE FOR PROBABILITY: MATH SHORT CUT TRICKS TO SOLVE PROBABILITY

There are two types of cases need to be discussed in the any PROBABILITY questions which are based on selection of a number among a defined set of numbers.

Let       P= Probability

            S= Special Case

            G= General Case

So        P (S) = Probability of Special Case

            P (G) = Probability of General Case

REMEMBER - ONE SIMPLE RULE: MATH SHORT CUT TRICKS TO SOLVE PROBABILITY

            P = P (S) / P (G)

Use Combination for making selection of numbers among a definite set of numbers.

Let       C = Combination

            n = Total numbers in a set

            r = selected numbers among set of numbers

Then

            ⁿC_r ⁼ n / r (n-r)

Where =  Factorial of a number (see example for clarity)

Note: For ‘AND’ = Use Addition of Probability & for ‘OR’ = Use Multiplication of Probability

Question 1:

A bag contains 4 Red and 5 Yellow & 6 Green balls. 3 balls are drawn (selected) at random.

a.       What is the probability that balls are drawn contains exactly two Red.

b.      What is the probability that balls are drawn contains exactly two Yellow.

Solution:

In this question:

Colors----- à                Red             &n bsp;    Yellow             Green                TOTAL (Set of Numbers = n)

Balls------ à                  4             &nbs p;        5             &nbs p;        6             &nbs p;        15

Read the statement and find the General Condition in question. In above question, the general condition is to draw 3 balls randomly which implies Probability of General Case i.e. n = 15, r = 3. So

                         P (G) = ¹⁵C­­­­­₃………&#823 0;………………..(i)

Now discuss the special cases P (S) as follows:

a.      What is the probability that balls are drawn contains exactly two Red.

In the above case 3 balls are drawn in which exactly 2 balls are drawn from Red and 1 ball is drawn except Red balls (i.e. Yellow + Green balls = 5 + 6 = 11) that means

·         from set of Red balls total = 4 exactly 2 ball are drawn i.e. n = 4, r = 2

It means ⁴C₂.

·         from Yellow + Green balls total = 5 + 6 = 11 only 1 ball is drawn i.e. n = 11, r = 1

It means ¹¹C₁.

Combine the Special cases which are discussed above as:

                         P (S) = ⁴C₂ * ¹¹C₁ …………………………&#823 0;(ii)

Then Probability P will become

                        P = P (S) / P (G)

                         Put these values from (i) & (ii), We get

                         P = ⁴C₂ * ¹¹C₁ / ¹⁵C­­­­­₃

Now                  ⁴C₂ = (4*3*2*1)/( 2*1)*(4-2)!= 6

                         ¹¹C₁ = 11/ 1*(11-1) = 11

So          &nbs p;         P (S) = ⁴C₂ * ¹¹C₁ / ¹⁵C­­­­­₃ = 6 * 11 = 66

&                      P(G) = ¹⁵C­­­­­₃ = (15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 /( 3*2*1) * (15-3) = 455

Thus                P = P (S) / P (G) = 66 / 455     Ans.

b.      What is the probability that balls are drawn contains exactly two Yellow.

P (S) = ⁵C₂ * ¹⁰C₁

            P (S) = 10 * 10 = 100

&         P (G) = ¹⁵C­­­­­₃ = 15*14*13 / 3*2 = 2730/3*2=6 or 2730/6 = 455

Thus                P = P (S) / P (G) = 100 / 455   Ans.

Do the practice of such questions by using ONE SIMPLE RULE Method from any set of question papers of (SMART LEARNING WITH B M GAUR - THE NEXT-GEN COACHING) and score cut-off marks in SSC IBPS or other competitive exams.

PROBABLITY - the other way

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1.   Probability:- The theoretical probability of an event E, written as P(E) is defined as.

                                         P(E)= Number of outcomes Favorable to E

                                                   Number of all possible outcomes of the experiment

                          Where we assume that the outcomes of the experiment are equally likely.

2.   The probability of a sure event (or certain event) is 1.

3.   The probability of an impossible event is 0.

4.   The probability of an Event E is number P (E) such that 0≤P(E)≤1.

5.   Elementary events:- An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1.

6.   For any event E,P(E)+P()=1, where stands for not E, E and  are called complementary event.

7.   Performing experiments:-

a.   Tossing a coin.

b.   Throwing a die.

c.   Drawing a card from deck of 52 cards.

8.   Sample space:-The set of all possible outcomes in an experiment is called sample space.

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1. The probability of getting bad egg in a lot of 400 is 0.035.Then find the no. of bad eggs in the lot.

               = 400*0.035 = 14        [ans.14]

2. Write the probability of a sure event.   

= The probability of a sure event (or certain event) is 1.  [ans.1]

3. What is the probability of an impossible event. 

= The probability of an impossible event is 0.  [ans.0]

4. When a dice is thrown, then find the probability of getting an odd number less than 3.   

             = Odd number in adice 1,3,5 =3  Odd number less than 3 is only 1        = 1/6

Answer: 1/6

5. A girl calculates that the probability of her winning the third prize in a lottery is 0.08.If 6000 tickets are sold, how many ticket has she brought.   

=Chances are 0.08 in 1 therefore chances in 6000 are for  = 0.08/1*6000           = 480

Answer: 480

6. What is probability that a non-leap year selected at random will contain 53 Sundays.                    

A non-leap year has 365 days that means 365/7 = 52 weeks and 1 day. The probability of having 53 Sunday  in a non-leap year is 1 out of 7

Answer: 1/7

7. A bag contains 40 balls out of which some are red, some are blue and remaining is black. If the probability of drawing a red ball is  and that of black ball is, then what is the no. of black ball?          

11/20+1/5+x/1=40

22+8+40x=40

40x=40-30

X=10

Answer:10

8. Two coins are tossed simultaneously. Find the probability of getting exactly one head.               

The probability is 1 out of two that is ½

Answer: ½

9. A card is drawn from a well suffled deck of 52 cards. Find the probability of getting an ace.         

There are 4 ace in the deck of 52 cards therefore probability is 4/52 or 1 out of 13

Answer: 1/13

10. In a lottery, there are 10 prizes and 25 blanks. Find the probability of getting a prize.         

 P(E)= Number of outcomes Favorable to  = n(E)

           Number of all possible outcomes of the experiment = n(S)

P(E) = 10/35  or 2/7

Answer: 2/7

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1. Find the probability that a no. selected at random from the number 3,4,5,6,………..25 is prime.

Total numbers 3 to 25 = 23

Prime numbers in the series = 1,3,7,11,13,17,19,23 = 8

P(E) = 8/23

Answer: 8/23

2. A bag contains 5 red,4 blue and 3 green balls. A ball is taken out of the bag at random. Find the probability that the selected ball is (a) of red colour (b) not of green colour.                                                 

Total number of balls = 12

No of red balls = 5

Probability of selection of red balls is 5/12

No of Green balls is 3 No of Not-Green balls = 12-3 or 9

Probability of selection of red balls is 9/12 or 3/4

Answer: 5/12 and ¾

3. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability of drawing

  (a) A face card   (b) card which is neither a king nor a red card                                                        

Total number of cards = 52

Total number of face cards in the set of 52 = 3*4 = 12

Probability of drawing of face cards is 12/52 or 3/13(a)

Total number kings cards = 4 ( out of 2 are red and 2 are black)

Total number of red cards = 26 ( out of which 2 are king )  therefore remaining red cards = 4+24 = 28

Number of card which is neither a king nor a red card = 52-28 = 24

Probability of card which is neither a king nor a red card 24/52 or 6/13(b)

Answer: 3/13, 6/13

4. A dice is thrown once. What is the probability of getting a number greater than 4?   

A dice has 6 sides , each side contains 1,2,3,4,5,6 points/numbers

There are 5 and 6 i.e. 2 number is greater then 4

P(E) is 2/6 or 1/3

Answer: 1/3

5. Two dice are thrown at the same time. Find the probability that the sum of two numbers appearing on the top of the dice is more than 9

A dice has 6 sides, each side contains 1, 2, 3, 4, 5, 6 points/numbers

Sum of two numbers must be more than 9 those two numbers are 5+6 (1),  6+5(2), 4+6(3) and 6+4(4) or n(E)=4

The set of all possible outcomes in an experiment is called sample space is. n(S)

How much time the pair of two dice may be thrown?  = 6×6 or  n(S) = 36

P(E) is n(E)=4 /  n(S) = 36 0r 1/6

Answer: 1/6

6. Two dice are thrown at the same time. Find the probability of getting different numbers on both dice.

 Throws = 6 Pairs

Probability of results

1= 1,1  1,2  1,3  1,4  1,5  1,6  Probability =5

2= 2.1 2,2  2,3  2,4  2,5  2,6 Probability =5

3= 3,1 3,2 3,3  3,4  3,5  3,6 Probability =5

4= 4,1 4,2  4,3  4,4  4,5  4,6 Probability =5

5= 5,1  5,2  5,3 5,4  5,5  5,6 Probability =5

6= 6,1 6,2 6,3 6,4 6,5 6,6 Probability =5

We can see that there of 5 Probable Events, out of 6 getting different numbers on both dice

 P(E) of getting different numbers on both dice = 30/36 = 5/6

Answer: 5/6

7. A coin is tossed two times. Find the probability of getting almost one head.

 A COIN HAS TWO SIDES Head and tail

For two tosses the possible outcomes are: H,H or H,T or T,H or T,T 

If it is tossed two times total experiments will be 4

Out of these, 3 have heads in so the probability is ¾

Answer: ¾

8. Cards with numbers 2 to 101 are placed in a box. A card selected at random from the box. Find the probability that the card which is selected has a number which is a perfect square.     

N(S) = 101-1=100

N(E) = 2,9,16,25,36,49,64,94,100  = 9

P(E) = 9/100

Answer: 9/100

9. Find the probability of getting the letter M in the word “MATHEMATICS”.          

n(S) = 11

 n(E) = 2

 n(E) = 2/11  = 11

Answer: 2/11

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1. Cards bearing numbers 3, 5,…………..,35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing (a)a prime number less than 15 (b)a number divisible by 3 and 5.         

 n(S)    = 35-1-2 = 17

(a) Prime number less then 15 = 1,3,7,11,13 = 5

 n(E) (a) = 5/17

(b) a number divisible by 3 and 5 = 15, 30 = 2

 n(E) (b) = 2/17

Answer: 5/17, 2/17

                    

2. Two dice are thrown at the same time. Find the probability of getting (a) same no. on the both side (b) different no. on both sides.                                                                                                                   

Outcomes for Two Dice

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

N(S) = 6 + 6 = 36

p(E) (a) same no. on the both side = 6/36 or 1/6

p(E) (b) different no. on both sides. = 30/36 = 5/6 

Answer:  1/6, 5/6                                                                                               

3. A child game has 8 triangles of which three are blue and rest are red and ten squares of which six are blue and rest are red. One piece is lost at random. Find the probability of that is (a) A square (b) A triangle of red colour.      

P(S) triangles = 8,  P(S) blue triangles = 3, p(S) red triangles = 5

P(S) square = 10, P(S) blue square = 6, p(S) red squares  = 4

One piece is lost at random out of 8+10 = 18

P(E) (a) A square = 10 of of 18 = 10/18  = 2/9

P(E) A triangle of red colour =  5 out of 18 = 5/18

Answer: 5/9, 8/18

4. Two dice are thrown simultaneously. What is the probability that:

(a) 5 will not come up either of them? (b) 5 will come up on at least one?  (C) 5 will come at both dice?    

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

Total possibilities = 6*6 = 36

Possibility of coming up of any one (1 to 6) in 1 throw is = 36-11=25

Possibility of not coming up of any one (1 to 6) in 1 throw is  = 6*2 = 12-1 =  11( in one pair a option come twice hence deducted)

(b) 5 will come up on at least one?  = pairs with 5 come only once = 11/36

(a) 5 will not come up either of them? = pairs without 5 = 36-11 = 25/36

(c) 5 will come at both dice? = only one probality = 1/36

Alternate

n(S)=36

n(E) 5 in all pairs = 6 + 6 -1= 11( in one pair 5 come twice hence deducted)

n(E) 5 in both dice = 1 P(E) = 1/36

n(E) 5 minimum once in = P(E) 5 in all pairs = 6 + 6=12 -1= 11( in one pair 5 come twice hence deducted) P(E)=11/36

n(E) 5 not come in either = 36 – 11= 25 P(E)=25/36

Answer:  25/36, 11/36, 1/36    

5. The king, queen and jack of clubs are removed from a deck of 52 playing cards and remaining cards are suffled. A card is drawn from the remaining cards. Find the probability of getting a card of (a)heart (b)queen (c)clubs   

N(S) = 52-3(removed 3 clubs)=49

N(E) (a)heart =   13 ,   P(E)  =   13/49

N(E) (b)queen = 4-1(removed 1 queen) = 3 ,   P(E)  = 3/49

N(E) (c)clubs    = 13-3(removed 3 clubs)=10  P(E)  = 10/49

Answer: 13/49, 3/49, 10/49

6. A game consists of tossing a one-rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e., 3 heads or three tails and looses otherwise. Calculate the probability that hanif will lose the game. 

ⁿ(S) = 1. H,T  2.  H,T  3. H,T  = HHH, HTH, TTT, THH = 4

Wins if TTT or HHH =1

ⁿ(E) Lose = HTH,  THH  and TTT or HHH = 3

p(E) lose = 3/4

Answer: 3/4

7. Cards bearing numbers 1,3,5,…………..,37 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing

(a) a prime number less than 15

ⁿ(S) = 37+1 = 38/2  = 19

ⁿ(E) a prime number less than 15 = 1, 3,7,11,13 = 5

p(E) = 5/19

Answer: 5/19

 (b)a number divisible by 3 and 5. 

ⁿ(E) a number divisible by 3 and 5 = 15, 30 = 2

p(E) = 2/19

Answer: 2/19

8. A dice has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and total score is recorded. (a) how many different scores are possible?  (b) what is the probability of getting a total of seven?  

ⁿ(S) = 0+1=1   0+6= 6  1+6=7, 0+0=0  1+1=2  6+6 = 12  = 5 scores

Answer: {a} 5 scores(0,1,2,6,7,12)  

ⁿ(S) 7 = 6+1, 6+1, 6+1 = 3

ⁿ(E) 7 = 6+1

p(E) 7 = 1/3

Answer:  (b) 1/3

Self Evaluation

1.   Three unbiased coins are tossed together. find the probability of getting

(i)          all heads

ⁿ(S) = T+T+T, T+T+H, T+H+H, T+H+T,      H+H+H, H+H+T. H+T+T, H+T+H = 8                                  

ⁿ(E) H+H+H =1                         

Answer:  1/8

(ii)         two heads

 ⁿ(S) = T+T+T, T+T+H, T+H+H, T+H+T,     
 H+H+H, H+H+T. H+T+T, H+T+H = 8                                                   
 ⁿ(E) =T+H+H, H+H+T, H+T+H = 3               

Answer:  3/8

(iii)        one heads 

ⁿ(S) = T+T+T, T+T+H, T+H+H, T+H+T,      H+H+H, H+H+T. H+T+T, H+T+H = 8                                  

ⁿ(E) =T+H+T, H+T+T, T+T+H = 3                                                                              

Answer:   3/8

(iv)        at least two heads     

ⁿ(S) = T+T+T, T+T+H, T+H+H, T+H+T,      H+H+H, H+H+T. H+T+T, H+T+H = 8                                  

ⁿ(E) = T+H+H,  H+H+H, H+H+T, H+T+H = 4            

= 4/8 or 1/2                                                              

Answer:  1/2

2. Two dice are thrown simultaneously; Find the probability of getting an even number as the sum.

ⁿ(S) = 6*6 = 36

ⁿ(E) even number = 3*6 =18

p(E) = 18/36 or 1/2

Answer:  1/2

3.   Cards marked with the number 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from the box . Find the probability that the number on the card is:

(i)          An even number

ⁿ(S) = 101-1=100

ⁿ(E) = 100/2 =50

p(E) = 50/100 or1/2

Answer:  ½

(ii)         A number less than 14

ⁿ(E) = 2 to 13 = 12

p(E) = 12/100 or 3/25

Answer:  3/25

(iii)        A number is perfect square                                                  

ⁿ(E) = 4,9,16,25,36,49,64,81,100 = 9

p(E) = 9/100

Answer:  9/100

(iv)       A prime number less than 20

ⁿ(E) = 1,3,5,7,11,13,17,19 = 8

p(E) = 8/100 or 2/25

Answer:  2/25

4. Out of the families having three children, a family is chosen random. Find the probability that the family has

1. Exactly one girl      

ⁿ(S) = g+g+g, g+g+b, , g+b+b,  g+b+g, b+b+b, b+b+g, b+g+g, b+g+b = 8

ⁿ(E) = g+b+b, b+b+g, b+g+b = 3

p(E) = 3/8

Answer:  9/8

                     2. At least one girl 

ⁿ(S) = g+g+g, g+g+b, , g+b+b,  g+b+g, b+b+b, b+b+g, b+g+g, b+g+b = 8

ⁿ(E) = g+g+g, g+g+b, , g+b+b,  g+b+g, b+b+g, b+g+g, b+g+b = 7

p(E) = 7/8

                     Answer:  7/8

                     3. At most one girl

ⁿ(S) = g+g+g, g+g+b, , g+b+b,  g+b+g, b+b+b, b+b+g, b+g+g, b+g+b = 8

ⁿ(E) =  g+b+b,  b+b+g, b+g+g, b+g+b = 4

p(E) = 4/8 or 1/2

                      Answer:  1/2

5.   Five card the ten, jack, queen, king, and ace of diamonds are well shuffled with their face down-ward. One card is picked up at random

(i)          What is the probability that the card is the queen?   

              ⁿ(S) = 5

              ⁿ(E) = 1

              p(E) = 1/5

              Answer:  1/5

(ii)         If the queen is drawn and put aside what is the probability that the second card picked up is

(a) an ace   

                       ⁿ(S) = 4

                        ⁿ(E) = 1

                      p(E) = 1/4

Answer:  1/4

 (b)      a queen                                                 

                       ⁿ(S) = 4

                        ⁿ(E) = 1=0

                       p(E) = 1=0/4

 Answer:  0

Probability of Tossing Three Coins

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Here we will learn how to find the probability of tossing three coins.

Let us take the experiment of tossing three coins simultaneously:

When we toss three coins simultaneously then the possible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; where H is denoted for head and T is denoted for tail.

Therefore, total numbers of outcome are 2³ = 8

The above explanation will help us to solve the problems on finding the probability of tossing three coins.

Worked-out problems on probability involving tossing or throwing or flipping three coins:

1. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times and no head appeared 50 times. 

If three coins are tossed simultaneously at random, find the probability of: 
(i) getting three heads, 
(ii) getting two heads, 
(iii) getting one head, 
(iv) getting no head

Solution: 

Total number of trials = 250.
Number of times three heads appeared = 70.
Number of times two heads appeared = 55.
Number of times one head appeared = 75.
Number of times no head appeared = 50.

In a random toss of 3 coins, let E₁, E₂, E₃ and E₄ be the events of getting three heads, two heads, one head and 0 head respectively. Then,

(i) getting three heads

P(getting three heads) = P(E₁)

      Number of times three heads appeared
⁼                   Total number of trials     

= 70/250
= 0.28

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(ii) getting two heads

P(getting two heads) = P(E₂)

      Number of times two heads appeared
⁼                 Total number of trials     

= 55/250
= 0.22

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(iii) getting one head

P(getting one head) = P(E₃)

      Number of times one head appeared
⁼                 Total number of trials     

= 75/250
= 0.30NextGen CAREER COACHING - SMART LEARNING WITH B M GAUR - SSC-BANK PO-CLERK - 9462900411

(iv) getting no head

P(getting no head) = P(E₄)

      Number of times on head appeared
⁼                 Total number of trials  

= 50/250
= 0.20

Note:

In tossing 3 coins simultaneously, the only possible outcomes are E₁, E₂, E₃, E₄ and P(E₁) + P(E₂) + P(E₃) + P(E₄)

= (0.28 + 0.22 + 0.30 + 0.20)
= 1
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2. When 3 unbiased coins are tossed once.

What is the probability of:
(i) getting all heads
(ii) getting two heads
(iii) getting one head
(iv) getting at least 1 head
(v) getting at least 2 heads
(vi) getting atmost 2 heads

Solution: 

In tossing three coins, the sample space is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
And, therefore, n(S) = 8.

(i) Getting all heads

Let E₁ = event of getting all heads. Then,
E₁ = {HHH}
and, therefore, n(E₁) = 1.
Therefore, P(getting all heads) = P(E₁) = n(E₁)/n(S) = 1/8.

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(ii) Getting two heads

Let E₂ = event of getting 2 heads. Then,
E₂ = {HHT, HTH, THH}
and, therefore, n(E₂) = 3.
Therefore, P(getting 2 heads) = P(E₂) = n(E₂)/n(S) = 3/8.

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(iii) Getting one head

Let E₃ = event of getting 1 head. Then,
E₃ = {HTT, THT, TTH} and, therefore,
n(E₃) = 3.
Therefore, P(getting 1 head) = P(E₃) = n(E₃)/n(S) = 3/8.

NextGen CAREER COACHING - SMART LEARNING WITH B M GAUR - SSC-BANK PO-CLERK - 9462900411

(iv) Getting at least 1 head

Let E₄ = event of getting at least 1 head. Then,
E₄ = {HTT, THT, TTH, HHT, HTH, THH, HHH}
and, therefore, n(E₄) = 7.
Therefore, P(getting at least 1 head) = P(E₄) = n(E₄)/n(S) = 7/8.

NextGen CAREER COACHING - SMART LEARNING WITH B M GAUR - SSC-BANK PO-CLERK - 9462900411

(v) Getting at least 2 heads

Let E₅ = event of getting at least 2 heads. Then,
E₅ = {HHT, HTH, THH, HHH}
and, therefore, n(E₅) = 4.
Therefore, P(getting at least 2 heads) = P(E₅) = n(E₅)/n(S) = 4/8 = 1/2.

NextGen CAREER COACHING - SMART LEARNING WITH B M GAUR - SSC-BANK PO-CLERK - 9462900411

(vi) Getting atmost 2 heads

Let E₆ = event of getting atmost 2 heads. Then,
E₆ = {HHT, HTH, HTT, THH, THT, TTH, TTT}
and, therefore, n(E₆) = 7.
Therefore, P(getting atmost 2 heads) = P(E₆) = n(E₆)/n(S) = 7/8

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These examples will help us to solve different types of problems based on probability of tossing three coins.

आपके अभ्यास के लिये प्रश्नपत्र डाउनलोड करने के लिये यहां क्लिक करें