Answer:
V)
VI)
Probabilit of at laast tro punchase an eleetrie dryep = 1- P(at most one of these austomays Punchases an electrie drygp. E 0.572 PC au five purchase gas) o 16 = PC) ,(say ) P(au fve puchase electrie) - o.005PCB), (say) of Rach type is punchasid) Plat least onL [P)P) +t 0.121 ニ 0.879
Let A be the event that the older pump will fail Let B be the event that the newer pump will fail Let An Bbe the event that both the pumps fail. The event that only older pump will fail is shown in Venn diagram as follows: Only A B A The probability of this event is given by: P(Only A) P(A)-P(AnB) Given that this value is 0.10 So, P(Only A) P(A)-P(AnB) 30.10-P(A)-P(AΩΒ) P(A)-P(AnB)= 0.10 P(A) 0.10+ P(AnB)
The event that only newer pump will fail is shown as shaded region in the following Venn diagram: Only B B A
The probability of this event is given by: P(Only B) P(B)-P(AB) Given that this value is 0.05. So, P(Only B) P(B)-P(AnB) 0.05 P(B)-P(AnB) P(B)-P(An B) = 0.05 P(B) 0.05+P(AnB)
The pumping system will fail on any given day if both pumps fail. The event that both pumps fail is equal to the event AnB Since the two pumps fail independently of one another, the probability of the event An B is given by: P(AnB) P(A) P(B) -(0.10+P(AnB)(0.o5+ P(AnB)
Let P(An B) x So, it can be rewrite as: P(AnB)=(0.10+P(AnB))x(0.05+P(AnB)) (0.10+x)(0.05+x) x0.005+0.10x+0.05x +x x0.005+0.15x+x 0.005-0.85x + x2 = 0 (x-0.8441) (x-0.0059)= 0 x -0.8441 or x 0.0059 Since the probability cannot be greater than the either of 0.05 or 0.10. To satisfy this condition the required probability value is P(AnB) = 0.0059|