Probability

Complete the following statements:

(i) Probability of an event E + Probability of the event ‘not E’ = ...............

(ii) The probability of an event that cannot happen is ............... . Such an event is called ...............

(iii) The probability of an event that is certain to happen is ............... . Such an event is called ............... .

(iv) The sum of the probabilities of all the elementary events of an experiment is ............... .

(v) The probability of an event is greater than or equal to ............... and less than or equal to ............... .

Answer

(i) 1

(ii) 0, impossible event

(iii) 1, sure or certain event

(iv) 1

(v) 0, 1

Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

(iv) A baby is born. It is a boy or a girl.

Answer

(i) It is not an equally likely outcome because the probability of it depends on many factors.

(ii) It is not an equally likely outcome because the probability depends on the ability of the player.

(iii) It is an equally likely outcome because the probability is equal in both cases, it can be either true or false.

(iv) It is an equally likely outcome because the probability is equal in both cases, it can either be a boy or a girl.

Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Answer

Tossing a coin is considered to be a fair way of deciding which team should get the ball at the beginning of a football game because it is an equally likely event, i.e, the probability of heads is $\dfrac{1}{2}$ which is the same as the probability of tails $\dfrac{1}{2}$.

Which of the following cannot be the probability of an event?

1. $\dfrac{2}{3}$

2. -1.5

3. 15%

4. 0.7

Answer

We know that,

Probability of an event A is :

0 ≤ P(A) ≤ 1.

Since, -1.5 is negative.

Hence, Option (2) is the correct option.

If P(E) = 0.05, what is the probability of ‘not E’?

Answer

We know that,

E and 'not E' are complementary events.

∴ P(E) + P(not E) = 1

⇒ 0.05 + P(not E) = 1

⇒ P(not E) = 1 - 0.05 = 0.95

Hence, P(not E) = 0.95

A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

(i) an orange flavoured candy?

(ii) a lemon flavoured candy?

Answer

(i) Since, there are no orange candies in the bag.

∴ No. of favourable outcomes for getting orange flavoured candy = 0.

Hence, probability of taking out an orange flavoured candy = 0.

(ii) As the bag has only lemon flavoured candy, so every time she takes out only lemon flavoured candy.

∴ The event is sure event.

Hence, probability of taking out lemon flavoured candy is 1.

It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Answer

Let A be the event that 2 students have same birthday and A' be the event that 2 students does not have same birthday.

So, A and A' are complementary events.

∴ P(A) + P(A') = 1

⇒ P(A) + 0.992 = 1

⇒ P(A) = 1 - 0.992 = 0.008

Hence, the probability that the 2 students have the same birthday = 0.008

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) red ?

(ii) not red ?

Answer

No. of possible outcomes = 8 (3 red + 5 black)

(i) No. of favourable outcomes (for drawing a red ball) = 3

P(drawing a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{8}$.

Hence, probability of drawing a red ball = $\dfrac{3}{8}$.

(ii) Since, there are only two coloured balls in the bag.

∴ P(drawing a red ball) + P(not drawing a red ball) = 1

⇒ $\dfrac{3}{8}$ + P(not drawing a red ball) = 1

⇒ P(not drawing a red ball) = 1 - $\dfrac{3}{8}$ = $\dfrac{5}{8}$

Hence, probability of not drawing a red ball = $\dfrac{5}{8}$.

A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be

(i) red?

(ii) white?

(iii) not green?

Answer

No. of possible outcomes = 17 (5 red + 8 white + 4 green)

(i) No. of favourable outcomes (for drawing a red marble) = 5

P(drawing a red marble)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{17}$.

Hence, probability of drawing a red marble = $\dfrac{5}{17}$.

(ii) No. of favourable outcomes (for drawing a white marble) = 8

P(drawing a white marble)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{17}$.

Hence, probability of drawing a white marble = $\dfrac{8}{17}$.

(iii) Since, there are only 3 coloured marbles in the bag.

∴ P(not drawing a green marble) = P(drawing a red or white marble)

No. of favourable outcomes (for drawing red or white marbles) = 13 (5 red + 8 white)

P(not drawing a green marble)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{13}{17}$.

Hence, probability of not drawing a green marble = $\dfrac{13}{17}$.

A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin

(i) will be a 50 p coin ?

(ii) will not be a ₹ 5 coin?

Answer

No. of possible outcomes = 180 (100 50p coins + 50 ₹ 1 coins + 20 ₹ 2 coins + 10 ₹ 5 coins)

(i) No. of favourable outcomes (of getting 50p coins) = 100.

P(getting a 50p coin) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{100}{180} = \dfrac{5}{9}$.

Hence, the probability that the coin will be a 50p coin = $\dfrac{5}{9}$.

(ii) No. of favourable outcomes (of getting ₹ 5 coins) = 10.

P(getting a ₹ 5 coin) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{10}{180} = \dfrac{1}{18}$.

We know that,

Event of getting a ₹ 5 coin and event of not getting a ₹ 5 coin are complementary events.

∴ P(getting a ₹ 5 coin) + P(not getting a ₹ 5 coin) = 1

⇒ $\dfrac{1}{18}$ + P(not getting a ₹ 5 coin) = 1

⇒ P(not getting a ₹ 5 coin) = 1 - $\dfrac{1}{18}$

⇒ P(not getting a ₹ 5 coin) = $\dfrac{17}{18}$

Hence, the probability that the coin will not be a ₹ 5 coin = $\dfrac{17}{18}$.

Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Answer

No. of possible outcomes = 13 (5 male fishes + 8 female fishes)

No. of favourable outcomes (for getting a male fish) = 5

P(getting a male fish) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{13}$.

Hence, probability of taking out a male fish = $\dfrac{5}{13}$.

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at

(i) 8 ?

(ii) an odd number?

(iii) a number greater than 2?

(iv) a number less than 9?

Answer

(i) There is only one eight in the dart.

No. of favourable outcomes (for pointing at eight) = 1

No. of possible outcomes = 8

P(pointing at eight) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{8}$.

Hence, the probability that arrow points at eight = $\dfrac{1}{8}$.

(ii) Odd numbers in the dart = 1, 3, 5, 7.

No. of favourable outcomes (for pointing at odd number) = 4

No. of possible outcomes = 8

P(pointing at odd number)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{8} = \dfrac{1}{2}$.

Hence, the probability that arrow points at odd number = $\dfrac{1}{2}$.

(iii) Numbers greater than 2 in the dart = 3, 4, 5, 6, 7, 8.

No. of favourable outcomes (for pointing at number greater than 2) = 6

No. of possible outcomes = 8

P(pointing at number greater than 2)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{8} = \dfrac{3}{4}$.

Hence, the probability that arrow points at number greater than 2 = $\dfrac{3}{4}$.

(iv) Numbers less than 9 in the dart = 1, 2, 3, 4, 5, 6, 7, 8.

No. of favourable outcomes (for pointing at number less than 9) = 8

No. of possible outcomes = 8

P(pointing at number less than 9)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{8} = 1$.

Hence, the probability that arrow points at number less than 9 = 1.

A die is thrown once. Find the probability of getting

(i) a prime number;

(ii) a number lying between 2 and 6;

(iii) an odd number.

Answer

In a single throw of dice.

Sample space = {1, 2, 3, 4, 5, 6}

No. of possible outcomes = 6

(i) Favourable outcomes (of getting a prime number) = 2, 3, 5.

No. of favourable outcomes = 3

P(getting a prime number)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$.

Hence, the probability of getting a prime number = $\dfrac{1}{2}$.

(ii) Numbers lying between 2 and 6 are 3, 4, 5.

No. of favourable outcomes = 3

P(getting a number lying between 2 and 6)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$.

Hence, the probability of getting a number lying between 2 and 6 = $\dfrac{1}{2}$.

(iii) Favourable outcomes (of getting an odd number) = 1, 3, 5.

No. of favourable outcomes = 3

P(getting an odd number)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$.

Hence, the probability of getting an odd number = $\dfrac{1}{2}$.

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

Answer

No. of possible outcomes = 52.

(i) There is one king of hearts and one king of diamonds in a deck.

No. of favourable outcomes (of getting a king of red colour) = 2

P(drawing a king of red colour)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}$.

Hence, the probability of drawing a king of red colour = $\dfrac{1}{26}$.

(ii) There are 12 face cards (3 of each suit) in a deck.

No. of favourable outcomes (for getting a face card) = 12

P(drawing a face card)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{12}{52} = \dfrac{3}{13}$.

Hence, the probability of drawing a face card = $\dfrac{3}{13}$.

(iii) There are 6 red face cards (3 of hearts and 3 of diamonds) in a deck.

No. of favourable outcomes (for getting a red face card) = 6

P(drawing a red face card)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{52} = \dfrac{3}{26}$.

Hence, the probability of drawing a red face card = $\dfrac{3}{26}$.

(iv) There is one jack of hearts.

No. of favourable outcomes (for getting a jack of hearts) = 1

P(drawing a jack of hearts)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{52}$.

Hence, the probability of drawing a jack of hearts = $\dfrac{1}{52}$.

(v) There are 13 spade cards in a deck.

No. of favourable outcomes(for getting a spade) = 13

P(drawing a spade)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{13}{52} =\dfrac{1}{4}$.

Hence, the probability of drawing a spade = $\dfrac{1}{4}$.

(vi) There is 1 queen of diamonds in a deck.

No. of favourable outcomes(for getting queen of diamonds) = 1

P(drawing a queen of diamonds)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{52}$.

Hence, the probability of drawing a queen of diamonds = $\dfrac{1}{52}$.

Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

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

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Answer

No. of possible outcomes = 5

(i) Out of cards,

No. of favourable outcomes(for getting a queen) = 1

P(drawing a queen) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{5}$.

Hence, the probability of drawing a queen = $\dfrac{1}{5}$.

(ii) Since, queen is drawn and put aside remaining cards = 4.

(a) No. of ace cards = 1

No. of possible outcomes = 4

No. of favourable outcomes (for drawing an ace) = 1

P(drawing an ace) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{4}$.

Hence, probability of drawing an ace = $\dfrac{1}{4}$.

(b) No. of queens left after putting a queen card aside = 0.

No. of possible outcomes = 4

No. of favourable outcomes (for drawing a queen) = 0

P(drawing a queen) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{0}{4}$ = 0.

Hence, probability of drawing a queen = 0.

12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Answer

No. of possible outcomes = 144 (12 defective pens + 132 good pens)

No. of favourable outcomes(for taking out good pen) = 132

P(drawing a good pen)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{132}{144} = \dfrac{11}{12}$.

Hence, probability of drawing a good pen = $\dfrac{11}{12}$.

(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?

Answer

No. of defective bulbs = 4

No. of favourable outcomes (of drawing a defective bulb) = 4

No. of possible outcomes = 20

P(drawing a defective bulb) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{20} = \dfrac{1}{5}$.

Hence, probability of drawing a defective bulb = $\dfrac{1}{5}$.

(ii) Since, one bulb is drawn and is not replaced.

∴ Remaining bulbs = 20 - 1 = 19.

∴ No. of possible outcomes = 19

Since, bulb drawn was not defective.

∴ Remaining good bulbs = 16 - 1 = 15

P(drawing a good bulb) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{15}{19}$.

Hence, probability of drawing a not defective bulb = $\dfrac{15}{19}$.

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5.

Answer

Since, discs are numbered from 1 to 90.

∴ No. of possible outcomes = 90

(i) There are 81 two digits numbers between 1 to 90.

∴ No. of favourable outcomes (of getting a disc with two-digit number) = 81

P(drawing a disc with two digit number)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{81}{90} = \dfrac{9}{10}$.

Hence, probability of drawing a disc with two digit number = $\dfrac{9}{10}$.

(ii) Perfect square numbers between 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81.

∴ No. of favourable outcomes (of getting a disc with perfect square) = 9

P(drawing a disc with perfect square)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{9}{90} = \dfrac{1}{10}$.

Hence, probability of drawing a disc with perfect square = $\dfrac{1}{10}$.

(iii) Numbers divisible by 5 between 1 to 90 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90.

∴ No. of favourable outcomes (of getting a disc with number divisible by 5) = 18

P(drawing a disc with number divisible by 5)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{18}{90} = \dfrac{1}{5}$.

Hence, probability of drawing a disc with number divisible by 5 = $\dfrac{1}{5}$.

A child has a die whose six faces show the letters as given below:

$\boxed{A} \space \boxed{B} \space \boxed{C} \space \boxed{D} \space \boxed{E} \space \boxed{A}$

The die is thrown once. What is the probability of getting

(i) A?

(ii) D?

Answer

There are six faces.

∴ No. of possible outcomes = 6

(i) No. of faces containing A = 6.

∴ No. of favourable outcomes = 2

P(of getting an A) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}$.

Hence, probability of getting an A = $\dfrac{1}{3}$.

(ii) No. of faces containing D = 1.

∴ No. of favourable outcomes = 1

P(of getting a D) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{6}$.

Hence, probability of getting a D = $\dfrac{1}{6}$.

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

Answer

Length of rectangular region = 3 m

Breadth of rectangular region = 2 m

Area of rectangular region = Length × Breadth

= 3 × 2 = 6 m².

Diameter of circular region = 1 m

Radius of circular region (r) = $\dfrac{1}{2}$ m

Area of circular region = πr²

= π × $\Big(\dfrac{1}{2}\Big)^2$

= $\dfrac{π}{4}$ m².

The probability that it will land inside the circle

= $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

= $\dfrac{\text{Area of circular region}}{\text{Area of the rectangular region}}$

= $\dfrac{\dfrac{π}{4}}{6} = \dfrac{π}{24}$.

Hence, the probability that die will land inside the circle is $\dfrac{π}{24}$.

A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it ?

(ii) She will not buy it ?

Answer

No. of possible outcomes = 144

(i) Nuri will buy pen if it is good.

No. of good pens = 144 - 20 = 124

∴ No. of favourable outcomes = 124

P(drawing a good pen)

= $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \dfrac{124}{144} = \dfrac{31}{36}$.

Hence, the probability that Nuri will buy the pen = $\dfrac{31}{36}$.

(ii) Buying a pen and not buying a pen are complementary events.

∴ P(buying a pen) + P(not buying a pen) = 1

⇒ $\dfrac{31}{36}$ + P(not buying a pen) = 1

⇒ P(not buying a pen) = 1 - $\dfrac{31}{36}$

⇒ P(not buying a pen) = $\dfrac{36 - 31}{36}$ = $\dfrac{5}{36}$

Hence, the probability of not buying a pen = $\dfrac{5}{36}$.

Complete the following table :

(ii) A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability $\dfrac{1}{11}$ . Do you agree with this argument? Justify your answer.

Answer

On tossing two dice.

Sample space = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}.

No. of possible outcomes = 36

Favourable outcomes for getting sum of 3 are {(1, 2), (2, 1)}

No. of favourable outcomes = 2

P(getting a sum of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{36}$.

Favourable outcomes for getting sum of 4 are {(1, 3), (2, 2), (3, 1)}

No. of favourable outcomes = 3

P(getting a sum of 4) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{36}$.

Favourable outcomes for getting sum of 5 are {(1, 4), (2, 3), (3, 2), (4, 1)}

No. of favourable outcomes = 4

P(getting a sum of 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{36}$.

Favourable outcomes for getting sum of 6 are {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

No. of favourable outcomes = 5

P(getting a sum of 6) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{36}$.

Favourable outcomes for getting sum of 7 are {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

No. of favourable outcomes = 6

P(getting a sum of 7) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{36}$.

Favourable outcomes for getting sum of 9 are {(3, 6), (4, 5), (5, 4), (6, 3)}

No. of favourable outcomes = 4

P(getting a sum of 9) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{36}$.

Favourable outcomes for getting sum of 10 are {(4, 6), (5, 5), (6, 4)}

No. of favourable outcomes = 3

P(getting a sum of 10) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{36}$.

Favourable outcomes for getting sum of 11 are {(5, 6), (6, 5)}

No. of favourable outcomes = 2

P(getting a sum of 11) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{36}$.

(ii) No, I don't agree because the argument is false. The eleven sums are not equally likely. Hence, each of them cannot have a probability of $\dfrac{1}{11}$. Also, in the above table we see that each sum has different probability.

Hence, the above argument is false.

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., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer

When one rupee coin is tossed three times.

Total possible outcomes are = {HHH, TTT, HTH, HHT, THH, THT, TTH, HTT}

∴ Number of possible outcomes = 8

∴ Number of favourable outcomes (to not get three heads or three tails) = 6.

P(Hanif will lose the game)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{8} = \dfrac{3}{4}$.

Hence, the probability that Hanif will lose the game is $\dfrac{3}{4}$.

A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

Answer

On throwing a dice twice.

Sample space = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}.

No. of possible outcomes = 36

(i) Favourable outcomes for 5 to not come up either time are {(1,1) (1,2) (1,3) (1,4) (1,6) (2,1) (2,2) (2,3) (2,4) (2,6) (3,1) (3,2) (3,3) (3,4) (3,6) (4,1) (4,2) (4,3) (4,4) (4,6) (6,1) (6,2) (6,3) (6,4) (6,6)}

No. of favourable outcomes = 25

P(that 5 will not come up either time)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{25}{36}$.

Hence, the probability that 5 will not come up either time is $\dfrac{25}{36}$.

(ii) Favourable outcomes for 5 to come up at least once are {(1,5) (2,5) (3,5) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,5)}

No. of favourable outcomes = 11

P(that 5 will come up at least once)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{11}{36}$.

Hence, the probability that 5 will come up at least once is $\dfrac{11}{36}$.

Which of the following arguments are correct and which are not correct? Give reasons for your answer.

(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is $\dfrac{1}{3}$.

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is $\dfrac{1}{2}$.

Answer

(i) The above statement is incorrect.

We can classify the outcomes like this but they are not then 'equally likely'. Reason is that ‘one of each’ can result in two ways — from a head on first coin and tail on the second coin or from a tail on the first coin and head on the second coin. This makes it twice as likely as two heads (or two tails).

(ii) The above statement is correct.

As, on tossing a die,

Sample space = {1, 2, 3, 4, 5, 6}.

Total possible outcomes = 6

Odd numbers = 3 {1, 3, 5}

Even numbers = 3 {2, 4, 6}.

P(odd numbers) = P(even numbers) = $\dfrac{3}{6} = \dfrac{1}{2}$.