Math Editor

Statistics Problems

Show that if A and B are independent events then B^C and A are independent.

independent events follow the multiplication law:

if A and B are independent then P(A∩B)=P(A)×P(B)

P(A∩BC)=P(A)−P(A∩B)=P(A)−P(A)×P(B)=P(A)×P(1−B)=P(A)×P(BC)

An unfair coin with P(T) = 1/3 is tossed three times.
Let A be the event of getting at exactly two heads and one tail.
B the event of getting tail in the second toss.

(1) Are A and B independent events?

(2) Evaluate P(A-B) and P(A/B).

Yes they are independent

P(H)=32​,P(T)=31​

S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}

A={HHT,HTH,THH}⟹P(A)=32​×32​×31​+32​×31​×32​+31​×32​×32​=94​

B={HTH,HTT,TTH,TTT}⟹P(B)=274​+272​+272​+271​=31​

A∩B={HTH}⟹P(A∩B)=274​

or because A and B are independent

P(A∩B)=P(A)×P(B)=94​×31​=274​

P(A−B)=P(A)−P(A∩B)=94​−274​=278​

or because A and B are independent

P(A−B)=P(A∩BC)=94​×32​=278​

P(A/B)=P(B)P(A∩B)​=1/34/27​=2712​=94​

or because A and B are independent

P(A/B)=P(A)=94​

Prove that the variance of Poisson distribution equals its mean.

the probability mass function for Poisson distribution is P(x)=x!λxe−λ​

the mean is μ=λ

μ=x=0∑∞​xx!λxe−λ​=x=1∑∞​xx(x−1)!λxe−λ​=e−λy=0∑∞​y!λy+1​=λe−λy=0∑∞​y!λy​=λe−λeλ=λ

the variance is v=μ

v=(x=0∑∞​x2x!λxe−λ​)−μ2=e−λ(x=1∑∞​x(x−1)!λx​)−μ2=e−λ(y=0∑∞​(y+1)y!λy+1​)−μ2=μ(μ+1)−μ2=μ

The number of accidents occur on a highway each day is a Poisson distribution with average 3

what is the probability that:

(1) At least one accident occurs today.

λ=3

P(x≥1)=1−P(x<1)=1−P(0)=1−0!30e−3​=95%

(2) 6 accidents occur in the next two days.

λ=6

P(6)=6!66e−6​=16%

The distribution function of a continuous random variable X is given by:

F(x)=⎩⎪⎨⎪⎧​2e2x​+c1​2kx2​+c2​c3​​x<00≤x<1x>1​

(1) Find the values of c₁, c₂, c₃ and k.

(2) Deduce PDF f(x)

F(−∞)=0⟹2e−∞​+c1​⟹c1​=0

F(∞)=1⟹c3​=1

F(0+)=F(0−)⟹21​=0+c2​⟹c2​=0.5

F(1+)=F(1−)⟹2k​+21​=1⟹k=1

F(x)=⎩⎪⎨⎪⎧​0.5e2x0.5x2+0.51​x<00≤x<1x>1​

f(x)=dxd​F(x)=⎩⎪⎨⎪⎧​e2xx0​x<00≤x<1x>1​

if X is a continuous uniform RV, show that E(x)V(x)​=6(b+a)(b−a)2​

f(x)=b−a1​

E(x)=2a+b​

V(x)=12(b−a)2​

prove it yourself!

The age of fans in a personal computer is exponentially distributed with average 1500 hr

a bulb is chosen at random.

(I) Deduce and sketch P. D. F f (x) and the distribution function F(x)

(2) Find the probability that it will work at least 2000 hr.

(3) Find the average such that 52.76% of bulbs will be destroyed before 3000 hr.

μ=λ1​=1500⟹λ=15001​

f(x)=λe−λx

F(x)=1−e−λx Deduce it yourself!

P(x≥2000)=1−P(x<2000)=1−F(2000)=e−2000/1500

F(3000)=0.5276=1−e−3000λ⟹λ=−3000ln(1−0.5267)​⟹μ=λ1​=4000

A four-sided die is tossed two times, let X and Y be random variables where

X: the sum of two numbers and Y: the minimum of two numbers.

Evaluate

(I) Joint probability mass function p(X, Y).

(2) cov(X, Y).

(3) p(x=5,Y<=2)

(4) p(x=2/Y=1)

E(X)=2×161​+3×162​+4×163​+5×164​+6×163​+7×162​+8×161​=5

E(Y)=1×167​+2×165​+3×163​+4×161​=1630​

E(X2)=4×161​+9×162​+16×163​+25×164​+36×163​+49×162​+64×161​=255​

E(Y2)=1×167​+4×165​+9×163​+16×161​=1670​

σ(X)=√E(X2)−E(X)2​=2√10​

σ(Y)=√E(Y2)−E(Y)2​=8√55​

E(X,Y)=2×161​+3×162​+4×162​+5×162​+8×161​+10×162​+12×162​+18×161​+21×162​+32×161​=885​

cov(X,Y)=E(X,Y)−E(X)E(Y)=885​−5×815​=1.25

correlation=σ(X)×σ(Y)cov(X,Y)​=2√10​×8√55​1.25​=0.8502 strong direct correlation

P(X=5,Y≤2)=162​+162​=41​

P(X=2/Y=1)=P(Y=1)P(X=2,Y=1)​=7/161/16​=71​

A Physics exam is made for 200 of Preparatory year students.

Their marks are normally distributed with mean = 12 and standard deviation = 2.5

assume that x denotes the mark of the student. Answer the following:

(1) Find the probability P(x≤14.5)

(2) If the excellent mark is more than 16, determine the number of the excellent students.

(3) Find the mark q ifϕ(q≤x≤14.5)=0.7719

P(x≤14.5)=ϕ(2.514.5−12​)=0.8413

Zexcellent​=2.516−12​=1.6

P(x≥16)=1−P(x<16)=1−ϕ(1.6)=0.0548

number of students =200×0.0548=11

Z of q must be negative

ϕ(2.514.5−12​)−(1−ϕ(Zq​))=0.7719

ϕ(Zq​)=0.7719−0.8431+1=0.9288

Zq​=1.47⟹−Zq​=−1.47⟹q=12−1.47×2.5=8.325

Two cards are drawn at random from a 52 card deck (without repiacement).

Find the probability of obtaining

I- Both cards are kings 2- Exactly one king 3. None of them is a king

1)524​×513​=2211​

2)524​×5148​+5248​×514​=22132​

3)5248​×5147​=221188​

1)52C24C2​=2211​

2)52C24C1×48C1​=22132​

3)52C248C2​=221188​

A box contains 4 white, 6 red,7 black and 3 green balls.

We draw 8 balls from the box (with replacement).

Find the probability of obtaining:

(1) two balls of each color

(2) One white ball only in the last drawing.

(3) two red balls in the third and fifth drawings

1)2!×2!×2!×2!8!​×(204​)2×(206​)2×(207​)2×(203​)2

2)(2016​)7×204​

3)2!×6!8!​×(206​)2×(2014​)6

A bag contains 5 fair coins and 10 unfair coins have 𝑃(𝐻) = 0.8.

A coin is picked at random and tossed 6 times.

Let 𝑩 the event of getting 4 heads out of 6 tosses.

Evaluate: 1- 𝑷(𝑭𝒂𝒊𝒓/𝑩). 2- 𝑷(𝑼𝒏𝒇𝒂𝒊𝒓/𝑩)

P(Fair)=31​,P(H/Fair)=21​,P(T/Fair)=21​

P(Unfair)=32​,P(H/Unfair)=108​,P(T/Unfair)=102​

P(B)=P(B/Fair)+P(B/Unfair)=1/3×6C4×0.54×0.52+2/3×6C4×0.84×0.22

P(Fair/B)=P(B)P(Fair∩B)​=1/3×6C4×0.54×0.52+2/3×6C4×0.84×0.221/3×6C4×0.54×0.52​

P(Unair/B)=P(B)P(Unair∩B)​=1/3×6C4×0.54×0.52+2/3×6C4×0.84×0.222/3×6C4×0.84×0.22​

An exponential distribution with density function 𝑓(𝑥) = 𝜆𝑒^−0.1𝑥 of a C.R.V

1) Find its mean

2) Deduce and draw the cumulative distribution 𝐹(𝑥).

μ=λ1​=0.11​=10

F(x)=1−e−0.1x

If X is a discrete random variable with a commutative distribution

F(X)=61​(x+k),x=1,2,3,4

(l) Find the value of k.

(2) Evaluate and sketch the probability mass function P(x) and F(x).

(3) Eva1uate E(6 - 2x), V(6 - 2x) and P(1.5 <= x <= 3.5).

61​((1+k)+(2+k)+(3+k)+(4+k))=1⟹k=−1

F(X)=61​(x−1)

F(1)=0,F(2)=61​,F(3)=62​,F(4)=63​

P(1)=61​,P(2)=61​,P(3)=61​,P(4)=63​

P(x)={61​63​​x=1,2,3x=4​

E(x)=1×61​+2×61​+3×61​+4×63​=3