Probability - Questions and Answers | Study

Question 1
a. The probability of a scrabble having the letters in the word HEXAGON are:
P( H ) = 2
100
P(E) = 12
100
P(X) = 1
100
P(A) = 9
100
P(G) = 3
100
P(O) = 8
100
P(N) = 6
100
The probability of having the word HEXAGON will be:
=P(H) and P(E) and P(X) and P(A) and P(G) and P(O) and P(N)
= 2
100 × 12
100 × 1
100 × 9
100 × 3
100 × 8
100 × 6
100
=3.1104 × 10𝑒−10
b. The probability of a scrabble the letters are:
P(G) = 3
100
P(A) = 9
100
P(R) = 6
100
P(B) = 2
100
P(E) = 12
100
The probability of having the word GARBAGE will be:
= P(G)2 andP(A)2 and P(R) and P(B) and P(E)
= 3
100
2
× 9
100
2
× 6
100 × 2
100 × 12
100
= 1.049× 10−9
c. The probability of having the letters in APPLE are:
P(A) = 9
100
P(P) = 2
100
P(L) = 4
100
P(E) = 12
100
The probability of having the word APPLE is:

= P(A)× P(P)2 × P(L) × P(E)
= 9
100 × 2
100
2
× 4
100 × 12
100
= 1.728× 10−7
Question 2
1). A is the event x1 (pellet)
P(A) = 25
29
B is the event x2 (pills)
P(A) = 4
28 (since one pellet had been consumed by the rat)
2). 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 1
= 25
29 + 4
28 − 1
= 1
203
3). 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
= 25
29 + 4
28 − 1
203
= 1
4). 𝑃(𝐴 ∩ 𝐵) = 1
203
𝑃(𝐴) × 𝑃(𝐵) = 25
29 × 4
28
= 25
203
Since 1
203 ≠ 25
203 then they are not independent.

Question 3
1). Pr(Head) = 1
2 Pr(Tail) = 1
2
n-number of trials (tosses) = 2
we need Pr(x≥ 1) = Pr(𝑥 = 1) + Pr (𝑥 = 2)
= (𝑛
𝑥)𝑝𝑛
=(2
1) 0.52 + (2
2)0.52
= 3
4 or 0.75
2). Pr(Head) = 1
2 Pr(Tail) = 1
2
n=10
We need Pr(x≥ 5) = 1 − Pr(𝑥 < 5)
= 1 − [(10
0 ) 0.510 + (10
1 ) 0.510 + (10
3 ) 0.510 + (10
4 ) 0.510
= 1 −[(1 + 10 + 45 + 120 + 210) × 0.510]
= 1 − 0.37695
= 319
512 or 0.6230
3). Pr(Head) = 1
2 Pr(Tail) = 1
2
n=2
We need Pr(x=1)
= (2
1) 0.52
= 1
2 or 0.5
4). Pr(Head) = 1
2 Pr(Tail) = 1
2
n=10
We need Pr(x=5)

= (10
5 ) 0.510
= 63
256 or 0.2461
Question 4
N=40
Pr(sitting on a chair) = 0.5
Pr(missing a chair) = 0.5
1). Pr(some chair has atleast 2 people) = Pr(𝑥 ≥ 2)
Pr(𝑥 ≥ 2) = 1 − (Pr(𝑥 = 0) + Pr(𝑥 = 1))
= 1 −[(40
0 ) × 0.540 + (40
1 ) × 0.540]
= 1 − 3.7289 × 10−11
= 0.9999
2). Pr(some chair has atleast 3 people)= Pr(x≥ 3)
= 1 −[(40
0 ) × 0.540 + (40
1 ) × 0.540 × (40
2 ) × 0.540]
= 1 − 7.46694 × 10−10
= 0.9999
Question 5
1). Pr(R1 is even) = 500
1000 = 1
2
Pr(R2 is even) = 1
2
Pr(R1 and R2 are even)= 1
2 × 1
2
= 1
4

2).
Pr(atleast one is even) = Pr(R1 even and R2 odd) or Pr(R1 odd and R2 even) or Pr( both R1and R2 even)
= (1
2 × 1
2) + (
1
2 × 1
2) + (
1
2 × 1
2)
= 3
4
But from P(A/B)= 𝑃(𝐴∩𝐵)×𝑃(𝐴)
𝑃(𝐵)
Pr(both R1and R2 even)/atleast one is even)= Pr(both R1and R2 even)×Pr (both R1and R2 even)
Pr (both R1and R2 even)
=
1
4×1
4
3
4
= 1
12
3). Pr( atleast one of R1 and R2 is equal to 1000)
=Pr(R1 is 1000 and R2≠ 1000) 𝑜𝑟 Pr ( R1 is 1000 and R2 ≠
1000 𝑜𝑟 Pr (R1 and R2 = 1000)
= 1
1000× 999
1000 + 1
1000× 999
1000 + 1
1000× 1
1000
= 1.998 × 10−6
4). P(A/B)= 𝑃(𝐴∩𝐵)×𝑃(𝐴)
𝑃(𝐵)
Pr(R1 or R2 is equal to 1000/ atleast R1 or R2 is even) = [Pr(atleast R1 or R2 =1000)]2
Pr( R1 or R2 is even)
= 1.998 × 10−6
3
4
=5.323× 10−6
5). P(A/B)= 𝑃(𝐴∩𝐵)×𝑃(𝐴)
𝑃(𝐵)
Pr(atleast one of R1 or R2 ≠ 999/ atleast R1 or R2 is even) )
= Pr(𝑅1=999 𝑎𝑛𝑑 𝑅2≠999)𝑜𝑟 Pr (𝑅2=999 𝑎𝑛𝑑 𝑅1≠999
Pr( atleast R1 or R2 is even)
=
1
1000× 999
1000+ 1
1000× 999
1000
3
4
= 1.330× 10−6
Question 6

N=10 n=7
Number of samples without repetition implies sampling without replacement
samples = (𝑁
𝑛)
= (10
7 )
= 120 samples
1). Pr(Max{x1,…x7} occurs only once in {x1,…x7}) = 1
120
Since there exists one Maximum in each sample.
2). Pr(Median occurs only once in {x1,…x7}) = 1
120
Since there exists one Median for each sample.