PROBABILITY-Exercise 14.1
Maths - Exercise
Q1. 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 called __________________. 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
_______________ .
Solution
(i) Probability of an event E + Probability of the event ‘not E’ = 1. (ii) The probability of an event that cannot happen is called 0. Such an event is called an impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure event (or certain event).
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.
Q2. 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.
Solution
Only experiments (iii) and (iv) can reasonably be treated as having equally likely outcomes at this level.Concept recap Two outcomes are called equally likely when each outcome has the same chance of occurring, with no reason to prefer one over the other.
Case-wise discussion
(i) A driver attempts to start a car.The chance that the car starts depends on many factors (condition of the engine, fuel, battery, weather, maintenance, etc.).
There is no reason to believe “starts” and “does not start” have the same probability, so the outcomes are not equally likely.
(ii) A player attempts to shoot a basketball.
The chance of scoring depends on the player’s skill, distance from the basket, practice, pressure, and other conditions.
The probabilities of “score” and “miss” are usually different, so the outcomes are not equally likely.
(iii) A trial is made to answer a true–false question.
Mathematically, if nothing is known about the question and a student guesses, “right” and “wrong” can be treated as having the same chance (each 1/2).
Under this ideal assumption, the two outcomes are equally likely.
(iv) A baby is born.
In school‑level probability, it is often assumed that the chances of “boy” and “girl” are about the same.
With this simplifying assumption, “boy” and “girl” are treated as equally likely outcomes.
Q3. 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?
Solution
Tossing a coin is considered fair because each team gets an equal chance to win the toss.When a standard (unbiased) coin is tossed, there are only two possible outcomes: head or tail. For a fair coin, the probability of getting a head is \(\frac{1}{2}\) and the probability of getting a tail is also \(\frac{1}{2}\), so neither outcome is favoured.
If one team chooses heads and the other chooses tails, both teams therefore have the same chance (50%) of winning the toss, making the decision process impartial and acceptable to both sides.
Q4. Which of the following cannot be the probability of an event?
(A) \(\frac{2}{3}\)
(B)–1.5
(C) 15%
(D) 0.7
Solution
The value that cannot be the probability of an event is (B) –1.5.Reason
The probability of any event always lies between 0 and 1, including 0 and 1, so it can never be negative or greater than 1.
(A) \(\frac{2}{3}\)=0.666…, (C) 15%=\(\frac{15}{100}\)= 0.15
(D) 0.7 is between 0 and 1, so it can be a valid probability.
(B) –1.5 is negative, so it cannot represent a probability.
Q5. If P(E) = 0.05, what is the probability of ‘not E’?
Solution
Q6. 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?
Solution
Since the bag has only lemon‑flavoured candies, picking is an example of an impossible and a certain event.(i) Probability of taking out an orange-flavoured candy
There is no orange‑flavoured candy in the bag, so the number of favourable outcomes is 0.
Therefore, \[P(\text{orange flavoured candy})=0\]
(ii) Probability of taking out a lemon-flavoured candy
Every candy in the bag is lemon‑flavoured, so every possible outcome is favourable.
Therefore, \[P(\text{lemon flavoured candy})=1\]
Q7. 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?
Solution
Probability of 2 students not having the same birthday \(P(E)\)=0.992Therefore, Probability that the 2 students have the same birthday = \(P(\overline{E})\) is compementary, hence \[ \begin{aligned} P(E)+P(\overline{E})&=1\\ P(\overline{E})&=1-P(E)\\ P(\overline{E})&=10.992\\ &=0.008 \end{aligned} \]
Probability that the 2 students have the same birthday is 0.008
Q8. 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?
Solution
Total number of possible outcomes = 3 + 5 = 8(i) Total number of favourable outcomes of drawing a red ball = 3, hence \[ \begin{aligned} P(R)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{3}{8} \end{aligned} \] (ii) Total number of favourable outcomes of drawing a black ball = 5, hence \[ \begin{aligned} P(R)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{5}{8} \end{aligned} \]
Q9. 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?
Solution
Total number of possible outcome = 5+8+4 = 17(i) Possible outcome that red marble is taken out = 5
hence, \[ \begin{aligned} P(R)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{5}{17} \end{aligned} \] (ii) Possible outcome that white marble is taken out = 8
hence, \[ \begin{aligned} P(R)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{8}{17} \end{aligned} \] (iii) Possible outcome that marble is taken out is not green = 5+8 =13
hence, \[ \begin{aligned} P(\overline{G})&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{13}{17} \end{aligned} \]
Q10. 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?
Solution
Total possible outcome = 100+50+20+10=180(i) Favourable outcome that dropped coin is 0.50 paise coin = 100 \[\begin{aligned} P(0.50 paise)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{100}{180}\\\\ &=\dfrac{5}{9} \end{aligned} \]
Probability that the coin will be a 50 p coin is \(\frac{5}{9}\)
(ii) Favourable outcome that the dropped coin is not a ₹ 5 coin = 170 \[\begin{aligned} P(\text{₹5})&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{170}{180}\\\\ &=\dfrac{17}{18} \end{aligned} \]Probability that the coin will be a ₹5 coin is \(\frac{17}{18}\)
Q11. 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 Fig. 14.4). What is the probability that the fish taken out is a male fish?
Solution
Total possible outcome = 5+8=13Favourable outcome to take out male fish = 5
Therefore, \[\begin{aligned} P(M)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{5}{13} \end{aligned} \]
Probability that the fish taken out is a male fish \(\frac{5}{13}\)
Q12. 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 Fig. 14.5 ), 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?
Solution
Total possible outcome = 8(i) Favourable outcome when the arrow points to 8 is = 1
Therefore, \[\begin{aligned} P(8)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{1}{8} \end{aligned} \]
Probability that arrow will point to numbers 8 is \(\frac{1}{8}\)
(ii) Favourable outcome when the arrow points to odd numbers = 4Therefore, \[\begin{aligned} P(O)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{4}{8}\\\\ &=\dfrac{1}{2} \end{aligned} \]
Probability that arrow will point to odd numbers \(\frac{1}{2}\)
(iii) Favourable outcome when the arrow points to numbers greater than 2 is = 6Therefore, \[\begin{aligned} P(E)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{6}{8}\\\\ &=\dfrac{3}{4} \end{aligned} \]
Probability that arrow will point to numbers greater than 2 is \(\frac{3}{4}\)
(iii) Favourable outcome when the arrow points to numbers less than 9 is = 8Therefore, \[\begin{aligned} P(F)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{8}{8}\\\\ &=1 \end{aligned} \]
Probability that arrow will point to numbers less than 9 is 1
Q13. 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.
Solution
Total possible outcome is 6 with 1, 2, 3, 4, 5, 6(i) Favourable outcome to get a prime number (2,3,5) = 3
Therefore, \[\begin{aligned} P(E)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{3}{6}\\\\ &=\dfrac{1}{2}\\\\ \end{aligned} \]
Probability od getting a prime number is \(\frac{1}{2}\)
(ii) Possible outcome to get a number lying between 2 and 6 is (3, 4, 5) = 3Therefore, \[\begin{aligned} P(F)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{3}{6}\\\\ &=\dfrac{1}{2}\\\\ \end{aligned} \]
Probability of getting a number lying between 2 and 6 is \(\frac{1}{2}\)
(iii)Possible outcome that it is an odd number (1, 3, 5) = 3Therefore, \[\begin{aligned} P(G)&=\scriptsize\dfrac{\text{No. of favaourable outcome}}{\text{No. of all possible outcome}}\\\\ &=\dfrac{3}{6}\\\\ &=\dfrac{1}{2}\\\\ \end{aligned} \]
Probability od getting an odd number is \(\frac{1}{2}\)
Q14. 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
Solution
In a standard well-shuffled deck, there are \(52\) cards in all. So, the total number of possible outcomes is \(52\).
(i) A king of red colour can be either the king of hearts or the king of diamonds, so there are \(2\) favourable outcomes. Using \(P(E) = \scriptsize\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}\), \[\begin{aligned} P(E) &= \dfrac{2}{52} \\\\&= \dfrac{1}{26} \end{aligned}\]
(ii) Face cards are jacks, queens, and kings. Each suit has three face cards, and there are four suits, so the total number of face cards is \(12\). Therefore, \[\begin{aligned} P(E) &= \dfrac{12}{52} \\\\&= \dfrac{3}{13} \end{aligned}\]
(iii) Red face cards belong to hearts and diamonds. Each of these two red suits has three face cards (jack, queen, king), giving \(6\) favourable outcomes in all. Hence, \[\begin{aligned} P(E) &= \dfrac{6}{52} \\\\&= \dfrac{3}{26} \end{aligned}\]
(iv) The jack of hearts is a single specific card, so there is exactly \(1\) favourable outcome. Thus, \[\begin{aligned} P(E) &= \dfrac{1}{52} \end{aligned}\]
(v) All spade cards form one complete suit, and each suit contains \(13\) cards, so there are \(13\) favourable outcomes. Therefore, \[\begin{aligned} P(E) &= \dfrac{13}{52} \\\\&= \dfrac{1}{4} \end{aligned}\]
(vi) The queen of diamonds is again one particular card in the deck, so the number of favourable outcomes is \(1\). Hence, \[ P(E) = \dfrac{1}{52} \]
Q15. 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?
Solution
Total number of possible outcome =5(i) possible outcome that picked card is the queen =1
Using \(P(E) = \scriptsize\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}\), \[\begin{aligned} P(E) &= \dfrac{1}{5} \\\\&= \dfrac{1}{5} \end{aligned}\]
Probability that one queen is picked is \(\frac{1}{5}\)
(ii) (a) Queen is kept outside, therefore total possible outcome =4Favourable outcome that an ace is picked = 1
Therefore, \[\begin{aligned} P(E) &= \dfrac{1}{4} \\\\&= \dfrac{1}{4} \end{aligned}\]
Probability that one ace is picked is \(\frac{1}{4}\)
(b) Possible outcome that a queen is picked = 0 as queen is aready taken out from the cardsTherefore, \[\begin{aligned} P(E) &= \dfrac{0}{4} \\\\&= 0 \end{aligned}\]
Probability that one queen is picked is 0
Q16. 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.
Solution
Total possible outcome = 12+132 = 144
favourable outcome to take one good pen = 132
therefore,
Probability that the pen taken out a good one is \(\frac{11}{12}\)
Q17. (i) A lot of 20 bulbs contains 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?
Solution
(i) total possible outcome = 20
favourable outcome to get a defective bulb = 4
Therefore, the probability of getting a defective bulb =
Probability of getting defective bulb =\(\frac{1}{5}\)
(ii) The defective bulb is taken out and not replaced.
Total possible outcome = 19
Favourable outcome to get a bulb which is not defective = 15
Therefore, the probability of getting a non-defective bulb =
Probability of getting a non defective bulb is \(\frac{15}{19}\)
Q18. 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.
Solution
Total possible outcome = 144
(i) favourable outcome when Nuri will buy the pen = 144-20 = 124
Therefore, the probability that Nuri will buy the pen =
Probability that Nuri will buy the pen \(\frac{31}{36}\)
(ii) Probability that she will not buy $$\begin{aligned}P\left( E\right) +P\left( \overline{E}\right) &=1\\\\ P\left( \overline{E}\right) &=1-P\left( E\right) \\\\ &=1-\dfrac{31}{36}\\\\ &=\dfrac{36-31}{36}\\\\ &=\dfrac{5}{36}\end{aligned}$$Probabity that she will not buy =\(\frac{5}{36}\)
Q19. A child has a die whose six faces show the letters as given below:
\[\boxed{A}\;\boxed{B}\;\boxed{C}\;\boxed{D}\;\boxed{A}\;\]
Solution
Total possible outcome = 6
(i) Favourable outcome of getting A = 2
Therefore, the probability of getting A =
Probability of getting A = \(\frac{1}{3}\)
(ii) Favourable out come to get D is 1
therefore, probability of getting D
Probability of getting D =\(\frac{1}{6}\)
Q20. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1m?
Solution
Area of rectangular region \(= 3\times2 = 6\ m_2\)
Diameter of circle \(= 1\ m\)
Hence, Radius of circle = 1/2 = 0.5 m
Area of circle =
Probability that the die drops in the circle
\[P(E) = \dfrac{\text{Area of circle}}{\text{Area of Rectangle}}\]
Q21.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?
Solution
Total possible out come = 144
(i) favourable outcome when Nuri will buy the pen = 144-20 = 124
Therefore, the probability that Nuri will buy the pen =
(ii) Probability that she will not buy
$$\begin{aligned}P\left( E\right) +P\left( \overline{E}\right) &=1\\\\ P\left( \overline{E}\right) &=1-P\left( E\right) \\\\ &=1-\dfrac{31}{36}\\\\ &=\dfrac{36-31}{36}\\\\ &=\dfrac{5}{36}\end{aligned}$$Q22. Refer to Example 13. (i) Complete the following table:
| Event: ‘Sum on 2 dice’ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Probability | \(\dfrac{1}{36}\) | \(\color{orange}\dfrac{2}{36}\) | \(\color{orange}\dfrac{3}{36}\) | \(\color{orange}\dfrac{4}{36}\) | \(\color{orange}\dfrac{5}{36}\) | \(\color{orange}\dfrac{6}{36}\) | \(\dfrac{5}{36}\) | \(\color{orange}\dfrac{4}{36}\) | \(\color{orange}\dfrac{3}{36}\) | \(\color{orange}\dfrac{2}{36}\) | \(\dfrac{1}{36}\) |
Solution
Total possibe outcome = 36(i) Favourable outcome for sum on dice =3
(1, 2), (2, 1)
P(E) = 2/36 = 1/18
(ii) Favourable outcome for sum = 4
(2, 2) (1,3) (3, 1)
P(E) = 3/36
(iii) Favourable outcome for sum = 5
(1,4), (4,3), (3, 2), (4, 1)
P(E) = 4/36
(iv) Favourable outcome for sun of 6
(1,5), (2/4), (3,3), (4,2), (5, 1)
P(E) = 5/36
(v) Favourable outcome to get sum of 7
(1, 6), (2, 5), (3, 4), (4, 3), (5,2), (6, 1)
P(E) = 6/36
(vi) Favourable outcome to get sum of 8
(2,6), (3,5), (4,4), (5,3), (6,2)
P(E) = 5/36
(Vii) Favourable outcome to get sum of 9
(3/6), (4,5), (5,4), (6,3)
P(E) = 4/36
(viii) Favourable outcome to get sum of 10
(4,6), (5,5), (6,4)
P(E) = 3/36
(ix) Favourable outcome to get sum of 11
(5/6), (6,5)
P(E) = 2/36
(x) Favourable outcome to get sum of 12
(6,6)
P(E) = 1/36
Q23. 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
Solution
Total possibe outcome = 8
Favourable outcome to lose Hanif
(H, H, T), (H, T, H), (T, H, H,), (T, T, H), (T, H, T), (H, T, T) \(\Rightarrow\) 6
Therefore, probabity that Hanif will lose the game
Q24. 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?
Solution
Total possible outcome = 36
\[\scriptsize
\begin{array}{|c|c|}
\hline(1,1)&(1,2)&(1,3)&(1,4)&\color{orange}(1,5)&(1,6)\\\hline
\hline(2,1)&(2,2)&(2,3)&(2,4)&\color{orange}(2,5)&(2,6)\\\hline
\hline(3,1)&(3,2)&(3,3)&(3,4)&\color{orange}(3,5)&(3,6)\\\hline
\hline(4,1)&(4,2)&(4,3)&(4,4)&\color{orange}(4,5)&(4,6)\\\hline
\hline\color{orange}(5,1)&\color{orange}(5,2)&\color{orange}(5,3)&\color{orange}(5,4)&\color{orange}(5,5)&\color{orange}(5,6)\\\hline
\hline(6,1)&(6,2)&(6,3)&(6,4)&\color{orange}(6,5)&(6,6)\\\hline
\end{array}
\]
(i) Favourable outcome to come 5 = 11
Therefore, a favourable outcome when 5 will not come either time = 36-11=25
Hence
Probability that I will not come either time =
Q25. 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 \(\frac{1}{3}\) (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 \(\frac{1}{2}\)
Solution
(i) The first argument is not correct. The actual sample space when two coins are tossed simultaneously is {HH, HT, TH, TT}, so there are four equally likely outcomes, not three. The events “two heads”, “two tails”, and “one of each” is not a single outcome:“Two heads” corresponds only to HH.
“Two tails” corresponds only to TT.
“One of each” corresponds to both HT and TH.
Hence, the probabilities are
\[P(\text{two heads })= \dfrac{1}{4}\] \[P(\text{two tails })= \dfrac{1}{4}\] \[\begin{aligned}P(\text{One Each })&= \dfrac{2}{4}\\\\ &=\dfrac{1}{2}\end{aligned}\]
so they are not all equal to \(\frac{1}{3}\)
(ii) The second argument is correct. When a die is thrown once, the sample space is {1,2,3,4,5,6}, giving six equally likely outcomes.The odd numbers are 1,3,5 and the even numbers are 2,4,6, so each group contains three outcomes.
Therefore, the probability of getting an odd number is \[ \begin{aligned} P(\text{odd number})&=\dfrac{3}{6}\\\\ &=\dfrac{1}{3} \end{aligned} \]
