The Practical business statistics

Assignment 4
Name of the student
Name of the University

Answer to question 1:
Let H denotes head and T tail
Pr(H) = Pr(T) = 1/2
[a] Pr(A)
= Pr(4H 2T) + Pr(5H 1A) + Pr(6H)
= ₆C (1/2)⁴(1/2)² +₄ ₆C (1/2)⁵(1/2) + (1/2)₅ ⁶
= 22/64
= 11/32
[b] Pr(B)
= Pr(3H 3T)
= ₆C (1/2)³(1/2)³₃
= 20/64
= 10/32
[c] Pr(C)
= Pr(HHHHT_) + Pr(THHHHT) + Pr(_THHHH) + Pr(HHHHHT) + Pr(THHHHH) +
Pr(HHHHHH)
= (1/2)⁴(1/2) + (1/2)(1/2)⁴(1/2) + (1/2)(1/2)⁴ + (1/2)⁵(1/2) + (1/2)(1/2)⁵ + (1/2)⁶
= (1/32) + (1/64) + (1/32) + (1/64) + (1/64) + (1/64)
= 1/8
[d] If the coin comes up heads at least four times, the coin comes up tails at most twice.
It is impossible that the number of heads is equal to the number of tails.
Pr(A and B) = 0
Pr(A|B)
= Pr(A and B) / Pr(B)

= 0
[e] Pr(C and A)
= Pr(C)
= 1/8
P(C|A)
= Pr(C and A) / Pr(A)
= (1/8) / (11/32)
= 4/11
Answer to question 2:
Given A = d1 + d2 = 7
B = d1 + d2 = 5
C = A U B
Hence, P(A) = 6/36
P(B) = 4/36
P(C) = P(A U B) = P(A) + P(B) [since, A and B cannot occur together]
= 6/36 + 4/36 = 10/36 = 5/18
Hence, P(A|C) = P(A)/P(C) = (6/36) / (5/18)
= 3/5
Answer to question 3:
[a] P(choosing the right door at the first go).
= ¼
[b] Since, Monte Hall opens a door and showing a goat, if one door is chosen randomly from
three unopened door, then