Exercise 14.1
Exercise 14.1
1. A die 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.
Sol:
a) The possible outcomes are 1,2,3,4,5,6.
b) Yes, all the outcomes are equally likely since every event has an equal chance of occurrence or no event has priority to occur.
c) Turning up a composite number of possible outcomes = 4,6
number of possible outcomes = 2
Total outcomes = 1,2,3,4,5,6
Number of total outcomes = 6
Probability = $$\frac{No.of\:favourable\:outcomes}{Total\:no.of\:outcomes}$$
= $$\frac{2}{6}$$
= $$\frac{1}{3}$$
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.
Sol:
a) Head = 45 times; tails = 55 times
Total = 100
Probability = $$\frac{No.of\:favourable\:outcomes}{Total\:no.of\:outcomes}$$
Probability of getting head p(H) = $$\frac{45}{100}$$
Probability of getting tail p(T)= $$\frac{55}{100}$$
b) p(H) + p(T) = $$\frac{45}{100}$$ + $$\frac{55}{100}$$ = $$\frac{100}{100}$$ = 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?
Sol:
a) Red = 5 sectors; blue = 3 sectors; green = 3 sectors; yellow =1 sectors;
Total = 5 + 3 + 3 + 1= 12 sectors
pointer is more likely to stop at red.
b) Yellow; as only one sector is shaded in yellow.
c) Blue and green have equal chances; as they are shaded in an equal number of sectors.
d) No chance. since no sector is shaded in white.
e) No; as the experiment is a random experiment.
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.
Sol:
a) No, as they are not in equal number, they have different chances of occurrence.
b) Green marbles + Blue marbles + Red marbles + Yellow marbles = 5+3+2+2= 12
Probability = $$\frac{No.of\:favourable\:outcomes}{Total\:no.of\:outcomes}$$
p(green) = $$\frac{5}{12}$$
p(blue) = $$\frac{3}{12}$$ = $$\frac{1}{4}$$
p(red) = $$\frac{2}{12}$$ = $$\frac{1}{6}$$
p(yellow) = $$\frac{2}{12}$$ = $$\frac{1}{6}$$
c) p(green) + p(blue) + p(red) + p(yellow) = $$\frac{5}{12}$$ + $$\frac{3}{12}$$ + $$\frac{2}{12}$$ + $$\frac{2}{12}$$ = $$\frac{12}{2}$$
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
Sol:
Total letters = 26 (A........Z)
Probability = $$\frac{No.of\:favourable\:outcomes}{Total\:no.of\:outcomes}$$
a) vowels =5 (a,e,i,o,u)
p(vowels) = $$\frac{5}{26}$$
b) letter after P = 10 (Q,R,S,T,U,V,W,X,Y,Z)
Probability = $$\frac{10}{26}$$ = $$\frac{5}{13}$$
C) A vowel or a consonants = 26 (A........Z)
Probability = $$\frac{26}{26}$$ = 1
d) Not a vowel = 21[other than a,e,i,o,u ]
Probability = $$\frac{21}{26}$$
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.
Sol:
Total number of bags = 11
No.of bags with weight more than 5 kg = 7( 5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07)
Probability = $$\frac{No.of\:favourable\:outcomes}{Total\:no.of\:outcomes}$$
P(E) = $$\frac{7}{11}$$.
7. An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained is given in the following table:
Find the probabilities of the following events for a driver chosen at random from
the city:
(i) The driver being in the age group 18-29 years and having exactly 3 accidents in one year.
(ii) The driver being in the age group of 30-50 years and having one or more accidents in a year.
(iii) Having no accidents in the year.
Sol:
i) Total number of accidents = (440 + 160 + 110 + 61 + 35 + 505 + 125 + 60 + 22 + 18 + 360 + 45 + 35 + 15 + 9) = 2000
Probability = $$\frac{No.of\:favourable\:outcomes}{Total\:no.of\:outcomes}$$
P(E) = $$\frac{61}{2000}$$.
ii) Favourable outcomes = 125 + 60 + 22 + 18 =225
Total outcomes = 2000
P(E) = $$\frac{225}{2000}$$ = $$\frac{9}{80}$$.
iii) Favourable outcomes = 440 + 505 + 360 =1305
Total outcomes = 2000
P(E) = $$\frac{1305}{2000}$$ = $$\frac{261}{400}$$.
8. What is the probability that a randomly thrown dart hits the square board in shaded region (Take π =$$\frac{22}{7}$$ and express in percentage)?
Sol:
The radius of the circle r = 2 cm
Area of the circle A = $$\pi r^2$$ = $$\frac{22}{7}\:\times 2\:\times 2$$ = $$\frac{88}{7}\:cm^2$$
side of the square = $$2\:\times\:radius$$ = $$2\:\times 2$$ = 4 cm
Area of the square = $$s^2$$ = $$4\times 4$$ = 16 $$cm^2$$
$$\therefore\:$$ Area of the shaded region = Area of the square - Area of the circle
= 16 - $$\frac{88}{7}$$
= $$\frac{24}{7}$$
Probability = $$\frac{No.of\:favourable\:outcomes}{Total\:no.of\:outcomes}$$
P(E) = $$\frac{\frac{24}{7}}{16}$$ = $$\frac{3}{14}$$.
As percentage = $$\frac{3}{14}\:\times 100$$ %
= $$\frac{300}{14}$$ %
= 21.428 %.
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