ID : in-0-probability [] Class 0 Probability For more such worksheets visit www.edugain.com Answer t he quest ions () Rajesh is participating in a race. The probability that he will come f irst in the race is 0.5. The probability that he will come second in the race is 0.. The probability that he will come in 3rd is 0.6, and the probability that he will be th is 0.5. What is the probability that he will come in 2nd position or better in the race? (2) The numbers to 9 are written on 9 pieces of paper and dropped into a box. Three of them are drawn at random. What is the probability that the three pieces of paper picked have numbers that are in arithmetic progression? (3) There are 5 hats belonging to 5 students. The hats were put into a box, and each student pulls out an hat one af ter the other. What is the probability that each student gets his or her own hat? () There are a total of 9 chocolates - 3 each in the f lavors of mango, banana and grape. There are also children. If each child is allowed to choose their own f avorite f lavor, what is the probability that all of them will get f lavors of their choice? (5) A bag contains 6 red balls, 6 blue balls, and 3 green balls. Aditya draws 2 balls out of the bag. What is the probability that he gets a green ball and a blue ball? (6) Archana draws cards out of a deck of cards. What is the probability that she draws a 2, a 9, a Queen and a? (7) Surjeet and Shilpa each have a bag that contains one ball each of the colors black, red, pink, green and blue. Surjeet randomly selects a ball f rom his bag and puts it in Shilpa's bag. Then Shilpa randomly selects a ball f rom her bag and puts it in Surjeet's bag. What is the probability that af ter this the contents of the bag are the same as bef ore? (8) A poll is taken among 0000 people working in a town. The aim was to see what their annual salaries were. Annual Salary Number of people Less than Rs.0000 20 Rs.000 to Rs.75000 660 Rs.7500 to Rs.50000 020 Rs.5000 to Rs.250000 3280 More than Rs.250000 920 If you choose a person at random f rom this group, what is the probability that he or she earns more than Rs.75000 annually?
ID : in-0-probability [2] Choose correct answer(s) f rom given choice (9) A tyre manuf acturer keeps the record of how much distance the tyres manuf actured by the company travel bef ore f ailing. They f ind the f ollowing data Distance traveled in kilometers Number of failing tyres Less than 5000 60 5000 to 0000 60 000 to 20000 300 2000 to 0000 592 More than 0000 888 If Shyam buys a tyre f rom them, what is the probability that it will last more than 0000 kilometers? a. 763 2000 c. 780 2000 78 2000 d. 788 2000 (0) From a deck of cards, Sneha took out one card at random. What is the probability that she got a prime number? a. c. 7 d. 2 () 6 coins are tossed in parallel. What is the probability of getting at least one heads? a. 6 63 6 c. 2 6 d. 30 6 (2) Usha draws one card f rom a shuf f led deck of cards. What is the probability that she drew a heart or a 5? a. c. 2 d.
() What is the probability that a leap year will contain 53 Sundays and Mondays? ID : in-0-probability [3] a. 366 c. 2 366 d. 7 () Coin A is f lipped 3 times and coin B is f lipped 5 times. What is the probability that the number of heads obtained f rom f lipping the two coins is the same? a. 3 32 7 32 c. 9 32 d. 3 30 (5) Two dice are rolled. What is the probability that the two numbers add up to a prime number? a. 5 36 2 36 c. 7 36 d. 5 36 206 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : in-0-probability [] () 0.25 Step We are looking f or the probability that he will come 2nd or better in the race This is the probability that he will either win the race or be 2nd Since he can be f irst or second, the probabilities can be added The probability that he will be f irst is 0.5 The probability that he will be in second place is 0. Adding them, we get 0.5 + 0. = 0.25
(2) 27 ID : in-0-probability [5] 323 Step This is a little complicated, so f ollow caref ully For making the explanation and the equations simpler, think of the number on the pieces of paper in the f orm of (2n+) Here, we can see f rom the equation 2n+ = 9, so n=9 The probability of getting 3 numbers in an A.P by selecting 3 numbers randomly between and 9 is the ratio of - Number of ways we can get an A.P f rom 3 random numbers between to 9, and - Number of ways to select 3 random numbers between to 9 Step Let's look at the second part f irst. Three tickets can be drawn f rom (2n+) numbers is in [(2 x n) + ] C 3 ways i.e. Number of ways 3 tickets can be drawn = (2n+)(2n)(2n-) 3x2x Simplif ying this, we get the number of ways to draw 3 numbers between and 9 = n(n 2 -) 3 Here n = 9, so we can simplif y it as 969 Step 5 Now f or the ways we can get an A.P f rom 3 numbers bwetween Arithmetic Progressions of 3 numbers would be a sequence of 3 numbers that are separated by a common interval e.g.,2,3 or 3,5,7 etc. They are in the f orm (a, a+d, a + 2d), where a is an integer f rom to (9-2), and d is another integer So it's helpf ul to think of the solution in terms of this interval. So we'll think of all the sequences that have an interval, then sequences with interval 2, and so on Step 6 So what are the possible sequences with interval. They are (,2,3) (2,3,)... (2n-,2n,2n+) T here are theref ore 2n- such possible sequences Step 7 Similarly, let's look at A.P with interval 2 between the terms. They are (,3,5) (2,,6)
...<2n-3,2n-,2n+> T here are 2n-3 such possible sequences ID : in-0-probability [6] Step 8 We can generalize this to say that the number of such sequences with interval 'd' is (2n- (2d-)) Obviously the largest possible integer is d=n, with just one sequence (, n+, 2n+) Step 9 So the total number of such sequences is (2n-) + (2n-3) + (2n-5) +... + 5 + 3 + This is itself an AP with n terms and d=2 The sum of this sequence is n 2 [2 + (n-)2] Simplif ying, we get n 2 Here n=9, so this is 8 Step 0 So the probability is 27 323 (3) 20 Step We have 5 hats and 5 students. Let's f irst see in how many dif f erent ways can the hats be distributed among the students We know that 5 objects can be distributed in 5! = 5 x x...x = 20 ways Out of these 20 ways, there is only one distribution where each student got his or her own hat Step So the probability of each student getting his or her own hat = 20
() 26 ID : in-0-probability [7] 27 Step There are children. They each have some f avourite f lavor f rom among mango, banana and grape. Each child's f avourite could be any one of the 3 choices The possible combinations of f lavors they like = 3 x 3 x 3 x 3 = 8 Let's now consider the options where even one child, no matter what his or her f avourite f lavour is, does not get his or her choice Now there are 3 f lavors in each choice, and children. The only way f or a child not to get his or her f avourite is if all children choose the same f lavor. This is because if even one child chooses some other f lavor f rom the rest, the other 3 children could get their f avourite, no matter what they choose The cases where some child might not get his or her choice is theref ore when they all choose the same f lavor Since there are 3 f lavors, this can happen in 3 cases 3 So the probability that a child does not get his or her f lavor = 8 Step Theref ore the probability that all children get their choice = - 3 8 = 26 27 (5) 8 05 Step There are a total of 6+6+3 = 5 balls in the bag The number of ways to pick out 2 balls f rom a set of 5 balls is 5 C2 = 5 x (5-)/2 = 05 The number of ways to pick a green ball and a blue ball is obtained by multiplying the number of the balls of each of these colors. This is 3 x 6 = 8 Step The probability that he gets a green ball and a blue ball is theref ore 8 05
(6) 256 ID : in-0-probability [8] 270725 Step There are 2s in a deck of cards. At this point there are also cards in the deck There are 9s in a deck of cards. Now there are 5 cards remaining in the deck There are Queens in a deck of cards. Now there are 50 cards remaining in the deck Step There are s in a deck of cards. Now there are 9 cards remaining in the deck Step 5 You can select cards without replacement f rom a deck in C ways = x 5 x 50 x 9 x 3 x 2 x Step 6 x x x So the answer is x 5 x 50 x 9 x 3 x 2 x Step 7 Simplif ying we get 256 270725 (7) 3
(8) 9220 ID : in-0-probability [9] 0000 Step First we need to f ind the total number of people among whom the poll was taken. We add the number of people in the various sets 20 + 660 + 020 + 3280 + 920 = 0000. To f ind the probability that the the random person chosen has a salary of more than Rs.75000, we need to add the number of people who have salaries greater than this number. This is 020 + 3280 + 920 = 9220. Step So, 9220 people out of a total of 0000 earn an annual salary greater than Rs.75000. The probability that the randomly chosen person has a salary greater than Rs.75000 = 9220. 0000 (9) c. 780 2000 Step First we need to f ind the total number of tyres that are given here. We add the number of tyres 60 + 60 + 300 + 592 + 888 = 2000. To f ind the probability that the tyre Shyam purchased would last more than 0000 kilometers, we need to add the number of tyres that lasted more than 0000 kilometers. This is 300 + 592 + 888 = 780. Step The probability that the tyre lasts more than 0000 km = 780 2000.
(0) Step ID : in-0-probability [0] There are cards in a deck, with of each suits (diamonds, clubs, hearts and spades Each of the cards of a suits are the numbers to 9, then f ace cards, including the ace In the numbers to 9, only 2,3,5,7 are primes. So f our primes per suit, which gives a total of 6 prime cards Step So there are 6 possible prime cards out of the total. The probability that we pick a prime card is theref ore 6, which can be reduced to () 63 6 Step The number of possible outcomes when 6 coins are tossed is 2 6 To f ind out the probability of getting at least one heads, let's look at the outcomes where this is not true i.e. the number of outcomes where you do not have even one heads Obviously, this is the case where you have all the tosses giving tails There is only one case where you can get all tails So the probability of getting at least one heads = - 6 = 63 6
(2) Step ID : in-0-probability [] The card she drew is one of the f ollowing quit - spade, diamond, club, or heart, with all of them being equally likely (as there are the same number of each f ace) The probability that she drew a heart is theref ore,. There are dif f erent cards of each suit, so f our of each f ace in the deck of. So probability that she drew a 5 is =. We need to add these two probabilities since the question asked f or the probability of getting a heart OR a 5. This gives us + = 7. Step But - there is another thing to keep track of. In our list, we've counted the 5 of heart twice, - once as a heart in the f irst component of the sum above, - and second time as a 5 in the second component above. We should only count it once, so we remove one of the times we count it. The probability of a specif ic card being chosen is, so we subtract that to get the result. Step 5 The probability that she drew a heart or a 5 = 7 - =.
() ID : in-0-probability [2] d. 7 Step There are 366 days in a leap year If we divide 366 by 7 (since there are seven days in a week), we will get an answer of, with a remainder of 2 This means that a leap year will have Sundays, Mondays, Tuesdays, Wednesdays, T hursdays, Fridays and Saturdays. Apart f rom these there will be two other days. This means that there will be two weekdays that occur 53 times. T he two days could be (Sunday, Monday), (Monday, T uesday), (T uesday, Wednesday), (Wednesday, T hursday), (T hursday, Friday), (Friday, Saturday), or (Saturday, Sunday) - a total of seven combinations Step Out of these seven combinations, only one of them has Sunday but no Monday Step 5 So the probability of either of those two days being a Sunday is 7 () 7 32
(5) a. 5 36 ID : in-0-probability [] Step The two dice that are rolled can show any of these values Dice :, 2, 3,, 5, 6 Dice 2 :, 2, 3,, 5, 6 So we can get a total of 36 combinations between them (6 x 6) If we take one value f rom the list of possible values f rom each Dice, we get numbers ranging f rom 2 (when both Dice show ) to 2 (when both dice show 6). Let's enumerate the prime numbers between 2 and 2. They are 2, 3, 5, 7 and We need to see in how many ways we can get each of these values Let's put the value rolled by the dice as (x,y), where x is the value rolled by Dice, and y the value rolled by Dice 2-2: The only way to get this is when we roll (,). possibility - 3: We can get this by (,2) or (2,). 2 possibilities. - 5: We can get this by (2,3), (3,2), (,) or (,). possibilities. - 7: We can get this by (,6), (2,5), (3,), (,3), (5,2), (6,). 6 possibilities. - : We can get this by (5,6), or (6,5). 2 Possibilities This gives us a total of + 2 + + 6 + 2 = 5 possible ways to get a prime number So the probability of getting the two numbers add up to a prime is 5 36