Probability — CBSE Class 10 Maths Textbook Solutions

(i) $P(E)+P(\text{not }E)=1$P(E)+P(not E)=1, since an event and its complement always cover all possible outcomes.

(ii) The probability of an event that cannot happen is 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 (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, i.e. $0\le P(E)\le 1$0≤ P(E)≤ 1.

(i) Not equally likely. Whether the car starts depends on its condition (battery, fuel, engine health), not on pure chance -- the two outcomes are not equally probable.

(ii) Not equally likely. Whether the shot goes in depends on the player's skill and the situation, not on pure chance.

(iii) Equally likely. If the answer is a pure guess with no other information, 'right' and 'wrong' are equally probable (each has probability $\frac{1}{2}$1/2).

(iv) Equally likely. Biologically, a newborn baby is (for this idealised experiment) equally likely to be a boy or a girl.

So, experiments (iii) and (iv) have equally likely outcomes.

When a fair coin is tossed, there are only two possible outcomes -- head or tail -- and because the coin is symmetric ('unbiased'), each outcome is equally likely to occur, i.e. $P(\text{head})=P(\text{tail})=\frac{1}{2}$P(head)=P(tail)=1/2.

This means the result of an individual toss is completely unpredictable and does not favour either team. That is why tossing a coin is considered a fair way of deciding which team gets the ball first.

The probability of any event must satisfy $0\le P(E)\le 1$0≤ P(E)≤ 1. It can never be negative.

$\frac{2}{3}\approx0.67$2/30.67, $15\%=0.15$15\%=0.15, and $0.7$0.7 all lie between 0 and 1, so they are valid probabilities. But $-1.5$-1.5 is negative, so it cannot be a probability.

Answer: (B) $-1.5$-1.5.

Using $P(E)+P(\text{not }E)=1$P(E)+P(not E)=1:
$P(\text{not }E)=1-P(E)=1-0.05=0.95$P(not E)=1-P(E)=1-0.05=0.95
The probability of 'not E' is 0.95.

(i) Since the bag contains only lemon flavoured candies, there are 0 outcomes favourable to drawing an orange flavoured candy.
$P(\text{orange flavoured candy})=\frac{0}{\text{total candies}}=0$P(orange flavoured candy)=0/total candies=0
This is an impossible event.

(ii) Since every candy in the bag is lemon flavoured, drawing a lemon flavoured candy is certain.
$P(\text{lemon flavoured candy})=1$P(lemon flavoured candy)=1
This is a sure event.

Let $E$E be the event 'the 2 students have the same birthday' and $\bar{E}$E be the event 'the 2 students do not have the same birthday'.
$P(\bar{E})=0.992 \text{ (given)}$P(E)=0.992 (given)
Using $P(E)=1-P(\bar{E})$P(E)=1-P(E):
$P(E)=1-0.992=0.008$P(E)=1-0.992=0.008
The probability that the 2 students have the same birthday is 0.008.

Total number of balls $=3+5=8$=3+5=8, all equally likely to be drawn.

(i) Number of outcomes favourable to 'red' $=3$=3.
$P(\text{red})=\frac{3}{8}$P(red)=3/8

(ii) Number of outcomes favourable to 'not red' (i.e. black) $=5$=5.
$P(\text{not red})=\frac{5}{8}$P(not red)=5/8
(This also follows from $P(\text{not red})=1-P(\text{red})=1-\frac{3}{8}=\frac{5}{8}$P(not red)=1-P(red)=1-3/8=5/8.)

Total number of marbles $=5+8+4=17$=5+8+4=17, all equally likely to be drawn.

(i) $P(\text{red})=\frac{5}{17}$P(red)=5/17

(ii) $P(\text{white})=\frac{8}{17}$P(white)=8/17

(iii) Number of green marbles $=4$=4, so $P(\text{green})=\frac{4}{17}$P(green)=4/17.
$P(\text{not green})=1-P(\text{green})=1-\frac{4}{17}=\frac{13}{17}$P(not green)=1-P(green)=1-4/17=13/17

Total number of coins $=100+50+20+10=180$=100+50+20+10=180, all equally likely to fall out.

(i) Number of 50p coins $=100$=100.
$P(50\text{p coin})=\frac{100}{180}=\frac{5}{9}$P(50p coin)=100/180=5/9

(ii) Number of Rs 5 coins $=10$=10, so $P(\text{Rs 5 coin})=\frac{10}{180}=\frac{1}{18}$P(Rs 5 coin)=10/180=1/18.
$P(\text{not a Rs 5 coin})=1-\frac{1}{18}=\frac{17}{18}$P(not a Rs 5 coin)=1-1/18=17/18

Total number of fish $=5+8=13$=5+8=13, all equally likely to be taken out.

Number of outcomes favourable to 'male fish' $=5$=5.
$P(\text{male fish})=\frac{5}{13}$P(male fish)=5/13

Total number of equally likely outcomes $=8$=8 (the numbers 1 to 8).

(i) Favourable outcomes to '8' $=1$=1 (just the number 8).
$P(8)=\frac{1}{8}$P(8)=1/8

(ii) Odd numbers among 1-8: 1, 3, 5, 7 -- 4 outcomes.
$P(\text{odd number})=\frac{4}{8}=\frac{1}{2}$P(odd number)=4/8=1/2

(iii) Numbers greater than 2: 3, 4, 5, 6, 7, 8 -- 6 outcomes.
$P(\text{number}>2)=\frac{6}{8}=\frac{3}{4}$P(number>2)=6/8=3/4

(iv) All numbers 1 to 8 are less than 9 -- 8 outcomes.
$P(\text{number}<9)=\frac{8}{8}=1$P(number<9)=8/8=1

Total number of equally likely outcomes when a die is thrown $=6$=6 (numbers 1 to 6).

(i) Prime numbers among 1-6: 2, 3, 5 -- 3 outcomes.
$P(\text{prime number})=\frac{3}{6}=\frac{1}{2}$P(prime number)=3/6=1/2

(ii) Numbers strictly between 2 and 6: 3, 4, 5 -- 3 outcomes.
$P(\text{number between 2 and 6})=\frac{3}{6}=\frac{1}{2}$P(number between 2 and 6)=3/6=1/2

(iii) Odd numbers among 1-6: 1, 3, 5 -- 3 outcomes.
$P(\text{odd number})=\frac{3}{6}=\frac{1}{2}$P(odd number)=3/6=1/2

Total number of equally likely outcomes $=52$=52 (well-shuffling ensures this).

(i) Red kings: king of hearts, king of diamonds -- 2 outcomes.
$P(\text{king of red colour})=\frac{2}{52}=\frac{1}{26}$P(king of red colour)=2/52=1/26

(ii) Face cards (king, queen, jack in each of 4 suits) $=4\times3=12$=4×3=12 outcomes.
$P(\text{face card})=\frac{12}{52}=\frac{3}{13}$P(face card)=12/52=3/13

(iii) Red face cards (king, queen, jack of hearts and diamonds) $=2\times3=6$=2×3=6 outcomes.
$P(\text{red face card})=\frac{6}{52}=\frac{3}{26}$P(red face card)=6/52=3/26

(iv) Only one jack of hearts.
$P(\text{jack of hearts})=\frac{1}{52}$P(jack of hearts)=1/52

(v) Number of spades $=13$=13.
$P(\text{spade})=\frac{13}{52}=\frac{1}{4}$P(spade)=13/52=1/4

(vi) Only one queen of diamonds.
$P(\text{queen of diamonds})=\frac{1}{52}$P(queen of diamonds)=1/52

(i) There are 5 equally likely outcomes (ten, jack, queen, king, ace), and 1 of them is the queen.
$P(\text{queen})=\frac{1}{5}$P(queen)=1/5

(ii) After the queen is drawn and put aside, 4 cards remain: ten, jack, king, ace.

(a) Number of outcomes favourable to 'ace' $=1$=1 out of the remaining 4.
$P(\text{ace})=\frac{1}{4}$P(ace)=1/4

(b) The queen has already been removed, so there is no queen left among the remaining 4 cards.
$P(\text{queen})=\frac{0}{4}=0$P(queen)=0/4=0

Total number of pens $=12+132=144$=12+132=144, all equally likely to be drawn.

Number of good pens $=132$=132.
$P(\text{good pen})=\frac{132}{144}=\frac{11}{12}$P(good pen)=132/144=11/12

(i) Total bulbs $=20$=20, defective bulbs $=4$=4.
$P(\text{defective})=\frac{4}{20}=\frac{1}{5}$P(defective)=4/20=1/5

(ii) Since the bulb drawn in (i) was not defective and is not put back, 19 bulbs remain, still containing all 4 defective bulbs. So the number of good (non-defective) bulbs remaining $=19-4=15$=19-4=15.
$P(\text{not defective})=\frac{15}{19}$P(not defective)=15/19

Total number of discs $=90$=90, all equally likely to be drawn.

(i) Two-digit numbers from 1 to 90 are 10, 11, ..., 90, i.e. $90-10+1=81$90-10+1=81 numbers.
$P(\text{two-digit number})=\frac{81}{90}=\frac{9}{10}$P(two-digit number)=81/90=9/10

(ii) Perfect squares from 1 to 90: 1, 4, 9, 16, 25, 36, 49, 64, 81 -- 9 numbers.
$P(\text{perfect square})=\frac{9}{90}=\frac{1}{10}$P(perfect square)=9/90=1/10

(iii) Multiples of 5 from 1 to 90: 5, 10, 15, ..., 90, i.e. $\frac{90}{5}=18$90/5=18 numbers.
$P(\text{divisible by 5})=\frac{18}{90}=\frac{1}{5}$P(divisible by 5)=18/90=1/5

Total number of faces $=6$=6: A, B, C, D, E, A. Note that the letter A appears on 2 faces.

(i) Number of faces showing A $=2$=2.
$P(A)=\frac{2}{6}=\frac{1}{3}$P(A)=2/6=1/3

(ii) Number of faces showing D $=1$=1.
$P(D)=\frac{1}{6}$P(D)=1/6

Since this involves landing anywhere within a region (infinitely many outcomes), probability here is calculated as the ratio of favourable area to total area.

Area of the rectangular region $=3\times2=6\text{ m}^2$=3×2=6 m^2.

Area of the circle (diameter 1 m, so radius $=0.5$=0.5 m) $=\pi r^2=\pi(0.5)^2=\frac{\pi}{4}\text{ m}^2$=π r^2=π(0.5)^2=π/4 m^2.

$P(\text{landing inside the circle})=\frac{\text{Area of circle}}{\text{Area of rectangle}}=\frac{\pi/4}{6}=\frac{\pi}{24}$P(landing inside the circle)=Area of circle/Area of rectangle=π/4/6=π/24

Total pens $=144$=144, defective $=20$=20, so good pens $=144-20=124$=144-20=124.

(i) Nuri buys the pen only if it is good.
$P(\text{she will buy it})=\frac{124}{144}=\frac{31}{36}$P(she will buy it)=124/144=31/36

(ii) Nuri will not buy it if it is defective.
$P(\text{she will not buy it})=\frac{20}{144}=\frac{5}{36}$P(she will not buy it)=20/144=5/36
(Also follows from $1-\frac{31}{36}=\frac{5}{36}$1-31/36=5/36.)

(i) When two dice are thrown, the total number of equally likely outcomes is $6\times6=36$6×6=36. Counting the number of (blue, grey) pairs that give each sum:

(ii) No, this argument is not correct. The 11 sums (2 through 12) are not equally likely -- each of the 36 individual (blue, grey) outcome pairs is equally likely, but different sums are made up of different numbers of these pairs. For example, only 1 pair gives a sum of 2 (namely (1,1)), while 6 pairs give a sum of 7. So the probabilities of the sums are not all $\frac{1}{11}$1/11; they range from $\frac{1}{36}$1/36 to $\frac{6}{36}$6/36 as shown in the table above.

When a coin is tossed 3 times, the total number of equally likely outcomes is $2^3=8$2^3=8:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

Hanif wins if all three tosses are the same: HHH or TTT -- 2 outcomes.
$P(\text{win})=\frac{2}{8}=\frac{1}{4}$P(win)=2/8=1/4

Hanif loses otherwise.
$P(\text{lose})=1-P(\text{win})=1-\frac{1}{4}=\frac{3}{4}$P(lose)=1-P(win)=1-1/4=3/4

When a die is thrown twice, the total number of equally likely outcomes is $6\times6=36$6×6=36.

(i) '5 will not come up either time' means each throw shows one of {1, 2, 3, 4, 6} -- 5 choices for each throw.
Number of favourable outcomes $=5\times5=25$=5×5=25.
$P(\text{5 does not come up either time})=\frac{25}{36}$P(5 does not come up either time)=25/36

(ii) '5 comes up at least once' is the complement of (i).
$P(\text{5 comes up at least once})=1-\frac{25}{36}=\frac{11}{36}$P(5 comes up at least once)=1-25/36=11/36

(i) Incorrect. While we can describe the outcomes as 'two heads', 'two tails' or 'one of each', these three outcomes are not equally likely. The actual equally likely outcomes when two coins are tossed are HH, HT, TH, TT (4 outcomes). 'Two heads' corresponds to only HH (1 outcome), 'two tails' corresponds to only TT (1 outcome), but 'one of each' corresponds to both HT and TH (2 outcomes). So 'one of each' is twice as likely as 'two heads' or 'two tails' -- the three groupings do not each have probability $\frac{1}{3}$1/3. The correct probabilities are $P(\text{two heads})=\frac{1}{4}$P(two heads)=1/4, $P(\text{two tails})=\frac{1}{4}$P(two tails)=1/4, $P(\text{one of each})=\frac{2}{4}=\frac{1}{2}$P(one of each)=2/4=1/2.

(ii) Correct. When a die is thrown, the outcomes 1, 2, 3, 4, 5, 6 are all equally likely. Odd numbers (1, 3, 5) and even numbers (2, 4, 6) each account for exactly 3 of the 6 equally likely outcomes, so the two outcomes 'odd number' and 'even number' are themselves equally likely, each with probability $\frac{3}{6}=\frac{1}{2}$3/6=1/2.