Respuesta :
Answer:
(a) 0.56
(b) 0.06
(c) 0.94
Step-by-step explanation:
P(E1) = 0.7 and P(E2) = 0.8
(a) The probability that both bids are successful is given by the product of the probability of success of each bid:
[tex]P(E1\ and\ E2) = 0.7*0.8=0.56[/tex]
(b) The probability that neither bid is successful is given by the product of the probability of failure of each bid:
[tex]P(not\ E1\ and\ not\ E2)= (1-P(E1))*(1-P(E2))\\P(not\ E1\ and\ not\ E2)=0.3*0.2=0.06[/tex]
(c) The probability that the firm is successful in at least one of the two bids is given by the sum of the probability of success of each bid subtracted by the probability that both bids are successful:
[tex]P(E1\ or\ E2)=P(E1)+P(E2) - P(E1\ and\ E2)\\P(E1\ or\ E2)=0.7+0.8-0.56\\P(E1\ or\ E2)=0.94[/tex]
Using probability concepts, it is found that there is a:
a) 0.56 = 56% probability that both bids are successful.
b) 0.06 = 6% probability that neither bid is successful.
c) 0.94 = 94% probability that the firm is successful in at least one of the two bids.
Item a:
These two events are independent, hence, the probability of both is the multiplication of the probabilities of each, thus:
[tex]p = 0.7(0.8) = 0.56[/tex]
0.56 = 56% probability that both bids are successful.
Item b:
E1 has a 1 - 0.7 = 0.3 probability of being unsuccessful, while E2 has a 0.2 probability, hence:
[tex]p = 0.3(0.2) = 0.06[/tex]
0.06 = 6% probability that neither bid is successful.
Item c:
1 - 0.06 = 0.94
0.94 = 94% probability that the firm is successful in at least one of the two bids.
A similar problem is given at https://brainly.com/question/23855473
