Then what will be the Mathematical chance for getting white ball. P (W) = 5/8 Black Ball. White Ball. Total P(B) P(W) First Box Second Box

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1 Possibilities as numbers There are 3 black balls and 5 white balls in a box. Suppose we are taking a ball from the box without peeping into it, what is the chance of getting a black ball. There are 8 balls in total and 3 black balls. So chance for getting black ball is 3 out of 8. This can be expressed in mathematics symbols as P(B) = 3/8, where P(B) stands for probability of getting Black ball. Here B is an event occurring at a particular time. Then what will be the Mathematical chance for getting white ball. P (W) = 5/8 Black Ball White Ball Total P(B) P(W) First Box Second Box Third Box All three boxes Now we can try the problems in page. (1) A box contains 6 black and 4 white balls. If a ball is taken from it, what is the probability of it being black? And probability of it being white? Answer : Total balls in the box =10

2 Probability of being black, P(B) = = Probability of getting white Ball, P(W) = = (2) There are 3 red balls and and 7 green balls in a bag. 8 red and 7 green balls in another. i) What is the probability of getting a red ball from the first bag? ii) From the second bag? iii)if all the bags are putting in single bag, what is the probability of getting a red ball? Answer : i) Probability of getting a red ball from the first bag P(R)= ii) From the second bag P(R)= iii) If all put together getting, chance for getting a red ball P(R) = (3) One is asked to say a 2 digit number. What is the probability of being a perfect square? Answer : The two digit numbers are 10,11,12,...,99 in total (99-10)+1 = 90 Two digit perfect squares are 16,25,36,49,64,81 in total 6 So probability of getting a perfect square = = (4) Numbers from 1 to 50 are written in paper slips and put it in a box. A slip is to be drawn from it, but before doing so, one must make a guess about the number, either prime number or a multiple of 5. Which is better guess?why? Answer : The slip contains 1,2,3...,50 total 50 slips. The prime numbers below 50 are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47 in total 15. Multiples of 5 below 50 5,10,15,20,25,30,35,40,45,50 total 10

3 Probability of getting a slip of prime number P(N) = Probability of getting Multiple of 5 P(M)= So it is better to guess the getting slip as prime number. (5) A bag contains 3 red beads and 7 green beads. Another contains one red and one green more. The probability of getting a red from which bag is more? Answer : P(R) from first bag = P(R) from second box = = (3x3 < 1x 10 so 1/3 is greater ) The possibility of getting red from the second box is better. Geometrical Probability a) What is the probability of the yellow part come towards the arrow mark in this figure when the wheel spins and let it naturally stop? Answer : Total 8 parts and 3 yellow parts, so probability is P(Y) = b) Suppose you are making a dot in the rectangle shown here by closing the eyes. What is the probability of that dot lie in the red triangle? Answer : The area of the red triangle is that of the rectangle. So the probability of the dot lies on the triangle is. Now we can try the problems here

4 In each figure we want to find out a dot putting by closing the eyes lie on the green part. (1) Green square obtained by joining the mid points of a bigger square. Answer : Let a is one side of the bigger square. Then side of green square =a / 2 Area of bigger square = a 2 Area of green square =a 2 /2 So probability is P(G) = (a 2 /2 )/a 2 = 1/2 (2) A square with all vertices on a circle. Answer : Diameter of the circle is the diagonal of the square. Area of the circle = r 2 = Area of square = x = ½d 2. Probability of dot in the square =Area of square / Area of circle = Answer : Probability of getting the dot inside the circle.=

5 Area of circle = Area of square. Let diameter of circle = d Area of circle = r 2 = One side of the square = d Area of square = side x side =d 2 Answer: Probability of getting a dot inside the triangle = Area of triangle is half of area of Hexagon

6 Answer: (Using formula 1/3) Pairs A) Johny own a pair of Blue pants and three shirts, Red Green, Blue. In How many ways he can dress? What is the probability of wearing pants and shirt of same colour? Answer : ( Blue pant, Red shirt ), ( Blue pant, Green shirt ), ( Blue pant, Blue

7 shirt ) by first pant. ( Blue pant, Red shir t ), ( Blue pant, Green shirt ), ( Blue pant, Blue shirt ) by second pant. So total 6 ways. Probability of wearing Pants and shirt of same colour = = answer: Total pairs =9 Pairs of same colours =3 = Probability of different colours = = answer: Total number of pairs = 8 Number of pairs with sum of numbers odd =4 Probability of sum even =

8 Home work Answer : Sample space= = 11,12,13,21,22,23,31,32,33 i) Numbers with both digit same 11,22,33 = ii) Numbers with sum 4 13,31,22 = = Answer: Even at the beginning is better. First hand {1,2,3,4,5} Second Hand {1,2,3,4,5} Total 5 x5 =25 out comes. Sums 1+1=2 3+1=4 5+1=6 1+2=3 3+2=5 5+2=7 1+3=4 3+3=6 5+3=8

9 1+4=5 3+4=7 5+4=9 1+5=6 3+5=8 5+5=10 2+1=3 4+1=5 2+2=4 4+2=6 2+3=5 4+3=7 2+4=6 4+4=8 2+5=7 4+5=9 Probability of getting even sum = Probability of getting odd sum = More pairs A )There are 50 mangoes in a basket, 20 of which are unripe. Another basket contains 40 mangoes, with 15 unripe. If we take one mango from each basket, i)what is the probability of both being ripe? ii) What is the probability of getting at least one ripe mango? Answer: i) Total number of sample space =50 x 40 =2000 The first basket has = 30 ripe ones and the second, = 25. Each ripe mango from the first basket paired with a ripe mango. This can be done in 30 x 25 =750 ways. So, Probability of getting a ripe mango = = ii) Total number of pairs with only one ripe =30 x x 25 = =950 Total number of pairs with both ripe = 750 Total number of pairs with at least one ripe = =1700 So, the probability of getting at least one ripe mango is =

10 Answer : i) Class 10 A 10 B Boys Girls Total Total pairs =40 x 40 =1600. Number of pairs with both girls = 20 x 25 =500 = ii) Number of pairs with both boys= 20 x 15 =300 = iii) Number of pairs with one boy and one girl = 20 x x 20= =800 iv) Number of pairs with at least one boy = 20 x x 15 =

11 Answer: Total sample space =10,11,12,13,...99 i) Number with both digit same =11,22,33,44,55,66,77,88,99 ii) Numbers with first number larger 10,20,21,30,31,32,40,41,42,43,50,51,52,53,54,60,61,62,63,64,65,... Total numbers = =45 probability = iii) Probability first digit being smaller = = Answer: First die {1,2,3,4,5,6 } Second {1,2,3,4,5,6} Total pairs 6 x 6 =36 possible sums =2,3,4,5,6,7,8,9,10,11,12 pairs with sum 2 are (1,1) probability = pairs with sum 3 are (1,2), (2,1)

12 probability = etc =========================ends============================