the answer is a)20,000 b)12,000 c)11000 d)68,000 e)10,000 atuandomly, the that h

Question

the answer is a)20,000 b)12,000 c)11000 d)68,000 e)10,000

atuandomly, the that he or she is not in any of the language (b) udent is chosen randomly, what is the probability that he or she is taking exactly one inopr anguage class? (c) students are chosen randomly, what is the probability that at least is taking a language class? 13. A certain town with a population of 100,000 has 3 newspapers: I, II, and Ill. The proportions 8 percent l and II and of townspeople who read these papers are as follows: 10 percent 4 (The list tells lll: 1 percent II: 30 percent I and Ill: 2 percent lll: 5 percent people who us, for instance, that 8000 people read newspapers I and IL) (a) Find the number of I and Ill are read only one newspaper. (b) How many people read at least two newspapers? (c) lf paper morning papers and II is an evening paper, how many people read at least one moming people plus an evening paper? (d) How many people do not read any newspapers? (e) How many read only one morning paper and one

Explanation / Answer

given total population 100000

let A denote event that proportion of people's read 1st news paper

B denote event that proportion of people's read 2nd news paper

C denote event that proportion of people's read 3rd news paper

P(A) =0.10 P(A and B)=0.08 P(A and B and C) =0.01

P(B)=0.30 P(A and C) =0.02 P(C) =0.05 P(B and C) =0.04

a)
we have to find no. of peoples who read only one newspaper

so first we find proportion of peoples who read only one news paper

i.e. we have to find proportion of peoples who read only one out of 1st ,2nd,or 3 rd paper

we have to find

P(A and B' and C') or P(A' and BandC') or P(A' and B' and C')

=P(A and B' and C') + P(A' and BandC') + P(A' and B' and C')

=P(A) -P(A and B) -P(A and C) +P( A and B and C) +P(B) -P(A and B) -P(Band C) +P( A and B and C)+P(C) -P(A and C) -P(B and C) +P( A and B and C)

=0.1-0.08-0.02+0.01 +0.3 -0.08-0.04+0.01+0.05-0.02-0.04+0.01

=0.01+0.19=0.2

so number of peoples =0.2*100000=20000

b)

number of peoples who reads at least 2 news papers

first we find proportion of peoples who reads A and B or B and C,or A and C or A and B and C

then required Proportion is (A and B) or (B and C) or (A and C) or (A and B and C)

=P(A and B) -P(A and B and C) +P(B and C)-P( A and B and C) +P( A and C)-P( A and B and C) +P(A and B and C)

=0.08-0.01+0.04-0.01+0.02-0.01+0.01

=0.12

hence no. of peoples =100000*0.12 =12000

c)

we have to find proportion of peoples who studying at least 1 morning paper (i.e. A and C) and evening paper

i.e.

we have to find P( A or C and B)

so

P( A or C and B )=P(( A and B) or (B and C) )

=P(A and B) +P( B and C) -P(A and B and C)

=0.08+0.04-0.01 =0.11

hence no. of peoples =100000*0.11 =11000

d)

we have to find no. of peoples who do not read any paper

we first find the Propotion of peoles who read at least one paper

so

P( A or B or C) =P(A) +P(B) +P(C)-P(A and B)-P( B and C)-P( A and C) +P(A and B and C)

=0.1+0.3+0.05 -0.08-0.04-0.02+0.1

=0.32

so proportion of peoples who do not read any paper =1-0.32 =0.68

hence required no. of peoples =100000*0.68 =68000

e)

we have to find number of peoples who read in morning only one of A or C and in evening B

i.e.

P(( A and B and C') or (A' and B and C))

=P( A and B and C') +P( A' and B and C)

=P(A and B)-P( A and B and C) +P( B and C) -P( A and B and C)

=0.08-0.01+0.04-0.01 =0.1

hence no. of peoles =0.1*100000=10000

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