**1** **$\textbf{Complete the following statements: } \\$ (i) Probability of an event E + Probability of the event ‘not E’ = _______. $\\$ (ii) The probability of an event that cannot happen is _________. Such as event is called _________.$\\$ (iii) The probability of an event that is certain to happen is _________. Such as 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 _______.$\\$**

(i) 1 $\\$ (ii) 0, impossible event$\\$ (iii) 1, sure event or certain event$\\$ (iv) 1$\\$ (v) 0, 1$\\$

**2** **$\textbf{Which of the following experiments have equally likely outcomes? Explain.}\\$ (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.$\\$**

(i) It is not an equally likely event, as it depends on various factors such as whether the car will start or not. And factors for both the conditions are not the same. $\\$ (ii) It is not an equally likely event, as it depends on the player’s ability and there is no information given about that. $\\$ (iii) It is an equally likely event.$\\$ (iv) It is an equally likely event.$\\$

**3** **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?**

When we toss a coin, the possible outcomes are only two, head or tail, which are equally likely outcomes. Therefore, the result of an individual toss is completely unpredictable.

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

The probability of an event (E) is always greater than or equal to 0. Also, it is always less than or equal to one. This implies that the probability of an event cannot be negative or greater than 1. Therefore, out of these alternatives, -1.5 cannot be a probability of an event.$\\$ Hence, (B)

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

We know that, P($\bar{E}$) = 1 - P(E) $\\$ P($\bar{E}$) = 1 - 0.05 $\\$ = 0.95 $\\$ Therefore, the probability of ‘not E’ is 0.95.

**6** **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?**

(i) The bag contains lemon flavoured candies only. It does not contain any orange flavoured candies. This implies that every time, she will take out only lemon flavoured candies. Therefore, event that Malini will take out an orange flavoured candy is an impossible event. $\\$ Hence, P (an orange flavoured candy) = 0 $\\$ (ii)As the bag has lemon flavoured candies, Malini will take out only lemon flavoured candies. Therefore, event that Malini will take out a lemon flavoured candy is a sure event. $\\$ P (a lemon flavoured candy) = 1

**7** **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?**

Probability that two students are not having same birthday P ($\bar{E}$) = 0.992 $\\$ Probability that two students are having same birthday $\\$ P(E) = 1-P($\bar{E}$) $\\$ = 1-0.992 $\\$ = 0.008

**8** **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 $\\$ $\textbf{(i) red? (ii) not red?}$**

(i) Total number of balls in the bag = 8 $\\$ Probability of getting a red ball = $\frac{Number of favourable outcomes}{Number of total possible outcomes}$ $\\$ = $\frac{3}{8}$ $\\$ (ii) Probability of not getting red ball = 1- Probability of getting a red ball $\\$ =1 - $\frac{3}{8}$ $\\$ =$\frac{5}{8}$

**9** **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 $\\$ $\textbf{(i) red? (ii) white? (iii) not green?}$**

Total number of marbles = 5 + 8 + 4 = 17 $\\$ (i)Number of red marbles = 5 $\\$ Probability of getting a red marble = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ $\\$ = $\frac{5}{17}$ $\\$ $\textbf{(ii)}$ Number of white marbles = 8 $\\$ Probability of getting a red marble = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ $\\$ = $\frac{8}{17}$ $\\$ $\textbf{(iii)}$ Number of green marbles = 4 $\\$ Probability of getting a red marble = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ $\\$ = $\frac{8}{17}$ $\\$ $\textbf{(ii)}$ Number of white marbles = 8 $\\$ Probability of getting a red marble = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ $\\$ = $\frac{4}{17}$ $\\$ Probability of not getting a green marble = 1 - $\frac{4}{17}$ = $\frac{13}{17}$

**10** **A piggy bank contains a hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 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 Rs.5 coin?$\\$**

Total number of coins in a piggy bank = 100 + 50 + 20 + 10 = 180 $\\$ $\textbf{(i)}$ Number of 50 p coins = 100 $\\$ Probability of getting a 50 p coin = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ $\\$ = $\frac{100}{180}$ = $\frac{5}{9}$ $\\$ $\textbf{(ii)}$ Number of 5 p coins = 10 $\\$ Probability of getting a 50 p coin = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ $\\$ = $\frac{10}{180}$ = $\frac{1}{18}$ $\\$ Probability of not getting a Rs 5 coin = 1 - $\frac{1}{18}$ = $\frac{17}{18}$

**11** **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 (see the given figure). What is the probability that the fish taken out is a male fish?**

Total number of fishes in a tank $\\$ = Number of male fishes + Number of female fishes $\\$ = 5 + 8 = 13 $\\$ Probability of getting a male fish = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ $\\$ = $\frac{5}{13}$

**12** **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 (see the given figure), 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?$\\$**

Total number of possible outcomes = 8 $\\$ $\textbf{(i)}$Probability of getting 8 = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ $\\$ = $\frac{1}{8}$ $\\$ $\textbf{(ii)}$ Total number of odd numbers on spinner = 4 $\\$ Probability of getting an odd number = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ $\\$ = $\frac{4}{8}$ = $\frac{1}{2}$ $\\$ $\textbf{(iii)}$ The numbers greater than 2 are 3, 4, 5, 6, 7, and 8. $\\$ Therefore, total numbers greater than 2 = 6 $\\$ Probability of getting a number greater than 2 = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ $\\$ = $\frac{6}{8}$ = $\frac{3}{4}$ $\\$ $\textbf{(iv)}$ The numbers less than 9 are 1, 2, 3, 4, 6, 7, and 8. $\\$ Therefore, total numbers less than 9 = 8 $\\$ Probability of getting a number less than 9 = $\frac{8}{8}$ = 1 $\\$

**13** **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.**

$\textbf{(i)}$ Prime numbers on a dice are 2, 3, and 5.$\\$ Total prime numbers on a dice = 3 $\\$ Probability of getting a prime number = $\frac{3}{6}$ = $\frac{1}{2}$ $\\$ $\textbf{(ii)}$ Numbers lying between 2 and 6 = 3, 4, 5 Total numbers lying between 2 and 6 = 3 The probability of getting a number lying between 2 and 6 = $\frac{3}{6}$ = $\frac{1}{2}$ $\\$ $\textbf{(iii)}$Odd numbers on a dice = 1, 3, and 5 $\\$ Total odd numbers on a dice = 3 $\\$ The probability of getting an odd number = $\frac{3}{6}$ = $\frac{1}{2}$$\\$

**14** **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**

Total number of cards in a well-shuffled deck = 52 $\\$ $\textbf{(i)}$ Total number of kings of red colour = 2 $\\$ The probability of getting a king of red colour = $\frac{Number of favourable outcomes}{Number of total possible outcomes}$ = $\frac{2}{52}$ = $\frac{1}{26}$ $\\$ $\textbf{(ii)}$ Total number of face cards = 12 $\\$ The probability of getting a face card = $\frac{Number of favourable outcomes}{Number of total possible outcomes}$ = $\frac{12}{52}$ = $\frac{3}{13}$ $\\$ $\textbf{(iii)}$ Total number of red face cards = 6 $\\$ The probability of getting a red face card = $\frac{Number of favourable outcomes}{Number of total possible outcomes}$ = $\frac{6}{52}$ = $\frac{3}{26}$ $\\$ $\textbf{(iv)}$ Total number of Jack of hearts = 1 $\\$ The probability of getting a Jack of hearts = $\frac{Number of favourable outcomes}{Number of total possible outcomes}$ = $\frac{1}{52}$ $\\$ $\textbf{(v)}$ Total number of spade cards = 13 $\\$ The probability of getting a spade card = $\frac{Number of favourable outcomes}{Number of total possible outcomes}$ = $\frac{13}{52}$ = $\frac{1}{4}$ $\\$ $\textbf{(vi)}$ Total number of queen of diamonds = 1 $\\$ The probability of getting a queen of diamond = $\frac{Number of favourable outcomes}{Number of total possible outcomes}$ = $\frac{1}{52}$ $\\$

**15** **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?**

(i) Total number of cards = 5 $\\$ Total number of queens = 1 $\\$ The probability of getting a queen = $\frac{Number of favourable outcomes}{Number of total possible outcomes}$ = $\frac{1}{5}$ $\\$ (ii) When the queen is drawn and put aside, the total number of remaining cards will be 4. $\\$ (a) Total number of aces = 1 $\\$ The probability of getting an ace = $\frac{Number of favourable outcomes}{Number of total possible outcomes}$ = $\frac{1}{4}$ $\\$ (b) As queen is already drawn, therefore, the number of queens will be 0. $\\$ The probability of getting a queen = $\frac{0}{4}$ = 0

**16** **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.**

Total number of pens = 12 + 132 = 144 $\\$ Total number of good pens = 132 $\\$ The probability of getting a good pen = $\frac{Number of favourable outcomes}{Number of total possible outcomes}$ = $\frac{132}{144}$ = $\frac{11}{12}$

**17** **(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?**

$\textbf{(i)}$ Total number of bulbs = 20 $\\$ Total number of defective bulbs = 4 $\\$ The probability of getting a defective bulb = $\frac{Number of favourable outcomes}{Number of total ossible outcomes}$ = $\frac{4}{20} = \frac{1}{5}$ $\\$ $\textbf{(ii)}$ Remaining total number of bulbs = 19 $\\$ Remainign total number of non-defective bulbs = 16-1 = 15 $\\$ The probability of getting a not defective bulb = $\frac{15}{19}$

**18** **Given that E and F are events such that $P ( E )= 0.6, P ( F )= 0.3$ and, find $P ( E | F )$ and $P ( F | E ) .$**

It is given that $P ( E )= 0.6, P (F )= 0.3 ,$ and $P ( E \cap F )= 0.2$$\\$ $\implies P(E|F) =\dfrac{P(E\cap F)}{P(F)} =\dfrac{0.2}{0.3} =\dfrac{2}{3}\\ \implies P(F|E) =\dfrac{P(E\cap F)}{P(F)} =\dfrac{0.2}{0.6} =\dfrac{1}{3}$