apssc probability exercise-2
Exercise-2
Example-8. A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be (i) white? (ii) blue? (iii) red?
Solution :
Saying that a marble is drawn at random means all the marbles are equally likely to be drawn.
`\therefore`The number of possible outcomes = 9 { 3blue, 2white, 4red balls }
(i) E ⟶ Event of drawing a White ball The number of outcomes favourable to the event White ball = 2
So, P(E) =`\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
`\therefore`P(E)=`\frac{2}{9}`
Similarly, (ii)E ⟶ Event of drawing a Blue ball
P(E) ==`\frac{3}{9}`=`\frac{1}{3}`
Similarly, (iii)E ⟶ Event of drawing a Red ball
P(E) ==`\frac{4}{9}`
Note that P(White balls) + P(Blue balls) + P(Red balls) = 1.
Example-9. Harpreet tosses two different coins simultaneously (say, one is of 1 and other of 2). What is the probability that she gets at least one head?
Solution :
We write H for 'head' and T for 'tail'. When two coins are tossed simultaneously, the possible outcomes =4 { (H, H), (H, T), (T, H), (T, T)}
The outcomes favourable to the event E, 'at least one head' are (H, H), (H, T) and (T, H). So, the number of outcomes favourable to E is 3.
`therefore`P(E) =`\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
i.e., the probability that Harpreet gets at least one head =`\frac{3}{4}`
Example-10. (Not for examination) In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting?
Solution :
Here the possible outcomes are all the numbers between 0 and 2. This is the portion of the number line from 0 to 2
Let E be the event that 'the music is stopped within the first half-minute'.
The outcomes favourable to E are points on the number line from 0 to `\frac{1}{2}`
The distance from 0 to 2 is 2, while the distance from 0 to is `\frac{1}{2}` is `\frac{1}{2}`
Since all the outcomes are equally likely, we can argue that, of the total distance is 2 and the distance favourable to the event E is `\frac{1}{2}`
So, P(E) =`\frac{Distance favourabLe To the event E}{Total distance In which outcomes can lie}`
=`\frac{1/2}{2}`=`\frac{1}{4}`
Example-11. A missing helicopter is reported to have crashed somewhere in the rectangular region as shown in the figure. What is the probability that it crashed inside the lake shown in the figure?
Solution : The helicopter is equally likely to crash anywhere in the region.
Area of the entire region where the helicopter can crash = (4.5 × 9) km2 = 40.5 km2
Area of the lake = (2 × 3) km2 = 6 km2
`\therefore` P (helicopter crashed in the lake) =`\frac{measure of Needed region part}{measure of whoLe region}`
=`\frac{6}{40.5}`=`\frac{60}{405}`=`\frac{4}{27}`
Example-12. A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jhony, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that (i) it is acceptable to Jhony? (ii) it is acceptable to Sujatha?
Solution :
One shirt is drawn at random from the carton of 100 shirts. Therefore, there are 100 equally likely outcomes.
(i)The number of outcomes favourable to Jhony = 88 ( 12 shirts are defected)
`\therefore` P (shirt is acceptable to Jhony) = `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
=`\frac{88}{100}`= 0.88
(ii)The number of outcomes favourable to Sujatha = 88 + 8 = 96 (`\because`4 have major defects)
So, P (shirt is acceptable to Sujatha) =`\frac{96}{100}`= 0.96
Example-13. Two dice, one red and one white, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8 (ii) 13 (iii) less than or equal to 12?
Solution :
When the red dice shows '1', the white dice could show any one of the numbers 1, 2, 3, 4, 5, 6. The same is true when the red dice shows '2', '3', '4', '5' or '6'.
The possible outcomes of the experiment are total number of possible outcomes
(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)
So, the number of possible outcomes
n(S) = 6 × 6 = 36.
(i)The outcomes favourable to the event the sum of the two numbers is 8 denoted by E, are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
i.e., the number of outcomes favourable to E is n(E) = 5.
Hence, P(E) =`\frac{n(E)}{n(S)}`=`\frac{5}{36}`
(ii)As there is no outcome favourable to the event F, the sum of two numbers is 13,
So, P(F) =`\frac{0}{36}`=0
(iii)As all the outcomes are favourable to the event G, sum of two numbers is less than or equal to 12,
So, P(G) =`\frac{36}{36}`=1
Exercise-2
1. 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 = 8 {3 red balls, 5 black balls }
(i) E ⟶ event of getting Red ball
Number of favourable outcomes = 3 {3 red balls}
Probability, P(E) =`3/8 `
(ii) `\bar E` ⟶ event of getting not a red ball
P (`\bar E`) = `1- p(E) = 1-3/8=(8-3)/8 = 5/8`
2. 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:
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
Total number of possible outcomes = 17 {5 red marbles, 4 green marbles & 8 white marbles}
(i) R ⟶ event of getting red marble
Number of favourable outcomes = 5 {5 red marbles}
P (R) =`5/17`
(ii) W ⟶ event of getting white marble
Number of favourable outcomes = 8 {8 white balls}
P(W) =`8/17`
(iii) `\bar G` ⟶ event of getting not green marble
Number of favourable outcomes = 13 {`\because`17-4}
P(`\bar G`) =`13/17`
3. A Kiddy 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:
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
Total number of possible outcomes = 180 { hundred 50p coins, fifty₹1 coins, twenty₹2coins & ten₹5 coins}
(i) E ⟶ event of getting 50p coin
P(E)= `100/180 = 5/9`
(ii) `\bar F` ⟶ event of getting not ₹5 coin
Number of favourable outcomes= 170 {`\because`180-10}
P(`\barF`)=`170/180 = 17/18`
4. 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 figure). What is the probability that the fish taken
out is a male fish?
Solution:
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
Total number of possible outcomes = 13 {5 male fish & 8 female fish}
E ⟶ event of getting male fish
Number of favourable outcomes = 5 {5 male fish}
P (E) =`5/13`
5. 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 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?
Solution:
Total number of possible outcomes = 8 {1, 2, 3,4,5,6,7,8}
(i) Let E ⟶ event of pointing 8
Number of favourable outcomes = 1 {8}
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`=`1/8`
(ii) Let O ⟶ event of pointing at an odd number.
Number of favourable outcomes = 4 {1, 3, 5, 7}
P(O) =`4/8=1/2`
(iii) Let T ⟶ event of pointing at a number greater than 2
Number of favourable outcomes = 6 {3,4,5, 6, 7,8}
P(T) =`6/8=3/4`
(iv) Let N ⟶ event of pointing at a number less than 9
Number of favourable outcomes = 8 {1,2,3, 4,5, 6,7, 8}
P(N) =`8/8=1`
6. 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:
Total number of possible outcomes = 52 (52 cards)
(i) R ⟶ event of getting a king of red colour
Number of favourable outcomes = 2 {king of heart & king of diamond}
P(R)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
P(R)=`2/52=1/26`
(ii) F ⟶ event of getting face card
Number of favourable outcomes = 12 {4 J,4 K & 4 Q}
P(F) =`12/52=3/13`
iii) RF ⟶ event of getting red face card
Number of favourable outcomes = 6 { J,Q,K of hearts & diamonds}
P(RF) = `6/52=3/26`
(iv) JH ⟶ event of getting a jack of heart
Number of favourable outcomes = 1 { jacks of hearts}
P(JH) =`1/52`
(v) S ⟶ event of getting a spade
Number of favourable outcomes = 13 {13 spades}
P(E) =`13/52=1/4`
(vi) QD ⟶ event of getting queen of diamond
Number of favourable outcomes = 1 { jacks of hearts}
P(QD) =`1/52`
7. 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 outcomes = 5 {5 cards}
(i) Q ⟶ event of drawing queen
Number of favourable outcomes = 1 {1 queen card}
P(Q)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }=1/5`
(ii)aWhen queen is drawn and put aside, total number of remaining cards = 4
Total number of possible outcomes = 4
Number of favourable outcomes = 1 {1 ace card}
P(Ace) = `1/4`
(ii)bNumber of favourable outcomes = 0 {card already drawn and put aside}
P(queen) = `0/4=0`
`8. 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:
Number of good pens = 132
Number of defective pens = 12
Total number of possible outcomes = 12 + 132= 144 {total no of pens}
E ⟶ event of getting a good pen.
Number of favourable outcomes = 132 {132 good pens}
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
`\therefore P(E) = 132/144 = 11/12`
9. 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? Suppose the bulb drawn in previous case 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:
Total number of possible outcomes = 20 {20 bulbs}
(i)
E ⟶ be event of getting defective bulb.
Number of favourable outcomes = 4 {4 defective bulbs}
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
`\therefore P(E) = 4/20= 1/5`
(ii)
Bulb drawn in is not defective and is not replaced with remaining bulbs = 15 + 4 = 19 bulbs
Total number of possible outcomes = 19
F ⟶ be event of getting not defective
Number of favorable outcomes = 15 (15 good bulbs)
P(F) =`15/19`
10. 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 number of possible outcomes = 90 {1, 2, 3, … 90}
(i)
E ⟶ event of getting 2 digit number.
Number of favourable outcomes = 81 {10, 11, 12, …. 90}
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
P(E) =`81/90=9/10`
(ii)
P ⟶ event of getting a perfect square.
Number of favourable outcomes = 9 {1, 4, 9, 16, 25, 26, 49, 64, 81}
P(E)=`9/90=1/10`
(iii)
F ⟶ event of getting a number divisible by 5.
Number of favourable outcomes = 18 {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}
P(E) =`18/90=1/5`
11. Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1m?
Solution:
circle radius = `1/2` m
Area of circle = π×(1/2)2 = `22/7×1/4 = 11/14`m2
Area of rectangle = 3 × 2 = 6m2
P(E) =`\frac{ measure of seLected region part}{Measure of whoLe region}`
P(E) =`\frac{ Area Of CircLe}{Area of rectangLe}`
Probability that die will land inside the circle with diameter 1m = `\frac{11/14}{6}=11/84`
12. A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that (i) She will buy it? (ii) She will not buy it ?
Solution:
Number of good pens = 144 – 20 = 24
Number of detective pens = 20
Total number of possible outcomes = 144 {total pens}
(i)
E ⟶ event of buying pen
Number of favourable outcomes = 124 {124 good pens}
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
P(E) =`124/144=31/36`
(ii)
`\bar E` ⟶ event of not buying a pen
P(E) + P(`bar E` ) = 1
P(`bar E`) = 1 - P(E)
P(`bar E`) = ` 1 - 31/36 =5/36`
13. Two dice are rolled simultaneously and counts are added (i) complete the table given
below:
Solution:
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
In a throw of pair of dice, total number of possible outcomes = 36 (6 × 6) which are
(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)
Two ⟶ event of getting sum as 2
Number of favourable outcomes = 1 {(1, 1)}
P(Two) =`1/36`
Three ⟶ event of getting sum as 3
Number of favourable outcomes = 2 {(1, 2) (2, 1)}
P(Three) =`2/36`
Four ⟶ event of getting sum as 4
Number of favourable outcomes = 3 {(3, 1) (2, 2) (1, 3)}
P(Four) =`3/36`
Five ⟶ event of getting sum as 5
Number of favourable outcomes = 4 {(1, 4) (2, 3) (3, 2) (4, 1)}
P(Five) =`4/36`
Six ⟶ event of getting sum as 6
Number of favourable outcomes = 5 {(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)}
P(Six) =`5/36`
Seven ⟶ event of getting sum as 7
Number of favourable outcomes = 6 {(1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)}
P(Seven) =`6/36`
Eight ⟶ event of getting sum as 8
Number of favourable outcomes = 5 {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
P(Eight) =`5/36`
Nine ⟶ event of getting sum as 9
Number of favourable outcomes = 4 {(3, 6) (4, 5) (5, 4) (6, 3)}
P(Nine) =`4/36`
Ten ⟶ event of getting sum as 10
Number of favourable outcomes = 3 {(4, 6) (5, 5) (6, 4)}
P(Ten) =`3/36`
Eleven ⟶ event of getting sum as 11
Number of favourable outcomes = 2 {(5, 6) (6, 5)}
P(Eleven) =`2/36`
Twelve ⟶ event of getting sum as 12
Number of favourable outcomes = 1 {(6, 6)}
P(Twelve) =`1/36`
14. 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:When 3 coins are tossed together,
Total number of possible outcomes = 8 {HHH, HHT, HTH, HTT, THH,THT, TTH, TTT}
(i)
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
E ⟶ getting other than only tails and heads
Number of favourable outcomes = 6 { HHT, HTH, HTT, THH,THT, TTH}
P( Haneef lose the game)=P(E)=`6/8=3/4`
15. 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? [Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].
Solution:
In a throw of pair of dice, total no of possible outcomes = 36 (6 × 6) which are
(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)
P(E)= `\frac{Number of favourabLe outcomes}{Total Number of possibLe outcomes }`
(i)
E ⟶ event of 5 will not come up either time
Number of favourable outcomes = 25 {(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)}
Total number of possible outcomes = 36
P(E) =`25/36`
(ii)
E ⟶ event of getting a 5 at least once
Number of favorable outcomes = 11 {(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6),(1, 5), (2, 5) (3, 5) (4, 5) (6, 5) }
Total number of possible outcomes = 36
P(E) =`11/36`
❮ Exercise 1
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