Telangana SCERT Class 9 Math Solution Chapter 14 Probability Exercise 14.1

Telangana SCERT Solution Class IX (9) Math Chapter 14 Probability Exercise 14.1

(1) A dice has six faces numbered from 1 to 6. It is rolled and the number on the top face is noted. When this is treated as a random trial.

a) What are the possible outcomes ?

b) Are they equally likely? Why?

c) Find the probability of a composite number turning up on the top face.

Solution:

(a) No of possible outcomes of a dice are the total no of numbers is each face i.e. 1,2,3,4,5,6.

∴ No of outcomes as = 6

(b) Yet they are equally likely because when the dice is rolled for large no of times the probability of each outcomes to occur becomes almost equal.

(c) Composite numbers is any positive integer greeter then 1 and not prime.

∴ No of possible composite numbers in a dice arc 4,6

∴ No of possible outcomes = 2

Total no of outcomes in a dice is 6

∴ P (Composite no turning up on top face) = No of possible outcomes/Total no of outcomes

= 2/6

= 1/3

(2.) A coin is tossed 100 times and the following outcomes are recorded

 Head:45 times Tails:55 times from the experiment

a) Compute the probability of each outcomes.

b) Find the sum of probabilities of all outcomes

Solution:

Given,

A coin was tossed 100 times

Total no of heads = 45 times

Total no of tails = 55 times

(a) ∴P (heads) = No of outcomes of head/Total no on outcomes

= 45/100

= 9/20

∴P (tails) = No of outcomes of tails/Total no on outcomes

= 55/100

= 11/20

(b) Sum of probability of all outcomes = 9/20 + 11/20

= (9+11)/20

= 20/20

= 1

(3.) A spinner has four colours as shown in the figure. When we spin it once, find

a) At which colour, is the pointer more likely to stop?

b) At which colour, is the pointer less likely to stop?

c) At which colours, is the pointer equally likely to stop?

d) What is the chance the pointer will stop on white?

e) Is there any colour at which the pointer certainly stops?

Solution:

(4.) A bag contains five green marbles, three blue marbles, two red marbles, and two yellow marbles. One marble is drawn out randomly.

a) Are the four different colour outcomes equally likely? Explain.

b) Find the probability of drawing each colour marble i.e. , P(green), P(blue), P(red) and P(yellow)

c) Find the sum of their probabilities.

Solution:

A bag contains 5 green marbles, 3 blue marbles, 2 red marbles, 2 yellow marbles

∴ Total no of marbles = 5 + 3 + 2 + 2= 12

P (green) = Total possible outcomes of green/Total no of outcomes

= 5/12

Similarly,

P (blue marbles) = 3/12

P (red marbles) = 2/12

P (yellow marbles) = 2/12

(a) The probability of 4 different color is not likely since the probability of each color is not equal

(b) P (green) = 5/12

P (blue) = 3/12 = ¼

P (red) = 2/12 = 1/6

P (yellow) = 2/12 = 1/6

(5.) A letter is chosen from English alphabet. Find the probability of the letters being

a) A vowel

b) a letter that comes after P

c) A vowel or a consonant

d) Not a vowel

Solution:

Total no of alphabets in English language = 26

Vowels in English language = a,e,i,o,u

∴ Total no of vowels = 5

(a) P (vowels) = Total no of vowels/Total no of alphabets

= 5/26

(b) Letter that comes after P are Q, R, S, T, U, V, W, X, Y, X

∴ Total no of Letter after p is 10

∴ P (a letter that comes after P) = Total no of letter after P/Total no of alphabets

= 10/26

= 5/13

(c) Consonants are letter except for vowels

∴ No of consonants = No of alphabets – no of vowels

= 26 – 5

= 21

∴ P (a vowel of consonant) = (total no of vowels + consonants)/Total no of alphabets

= (21+5)/26

= 26/26

= 1

(d) P (Not a vowel or total no of consonant) = total no of not vowels/total no of alphabets

= (26 – 5)/26

= 21/26

(6) Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):

4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00

Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Solution:

Weight of each bag of flour marked 5 are

4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00

Total no of bags = 11

Total no of bags containing more than 5 kg = 7

P (more than 5 kg bag) = Total no of bags containing more than 5kg/total no of 5kg bags

= 7/11

(7) Solution: The following distribution table is of 2000 drive of various age picked at random

Age of Drivers (in years) Accidents in one year More than 3 accidents
0 1 2 3
18 – 29 440 160 110 61 35
30 – 50 505 125 60 22 18
Over 50 360 45 35 15 9

(i) Total no of accident data collected for 18-29 years old individual = 44 + 160 + 110 +61 +36

= 806

Total no of drives of age 18 – 27 years having exactly 3 accidents in one year = 61

∴ P (drivers being age group 18 – 29 years having 3 accidents)

= Total no of drives with 3 accidents/Total no of data collected

= 61/806

(ii) Total no of accident data collected for 30 – 50 years old individuals = 505 + 125 + 60 + 22 + 18

= 730

Total no of accident caused by drives of age 30 – 50

= 125 + 60 + 22 + 18

= 225

∴ P (drives of age 30 – 50 yrs having one or more accidents)

= Total no of accident/total no of data collected

= 225/730

= 45/146

(iii) Total no of data collected for all age group

= 806 + 730 + 360 + 45 + 35 + 15 + 9

= 2000

Total no of drives of all age group causing no accidents

= 440 + 505 + 360

= 1305

P (Having no accidents in a year) = Total no of accidents with no accidents/Total death collected of all drivers

= 1305/2000

= 261/400

∴ P (randomly thrown dart hits the square board in shaded region)

= area of the shaded region/Total area of the square

= (24/7)/16

= 3.43/16

Updated: October 5, 2021 — 4:31 pm

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