Sounds like a problem in binomial probability. What do you think?
Here the # of experiments is 6, so n = 6. The probability of a baby girl being born is 0.50.
Using my TI-83 Plus calculator:
binompdf(6, 0.5, 2) = 0.234
binompdf(6, 0.5, 3) = 0.313
binompdf(6, 0.5, 4) = 0.234
binompdf(6, 0.5, 5) = 0.094
binompdf(6, 0.5, 6) = 0.016
To get the prob. of at least 2 girls in 6 births, add up the 5 probabilities given above:
P(at least 2 girls in 6 births) = 0. ???
Answer:
a) 0.623 (62.3%)
b) 0.764 (76.4%)
c) 0.6 (60%)
Step-by-step explanation:
a) defining the event F= the product is favorable , then the probability is
P(F) = probability that the study was sponsored by food industry * probability that the product is favorable given that it was sponsored + probability that the study had no corporate ties * probability that the product is favorable given that it was not sponsored = 0.70 * 0.68 + 0.30 * 0.49 = 0.623
b) for conditional probability we use the theorem of Bayes , then defining the event S= the study was sponsored by the food industry , we have:
P(S/F)=P(S∩F)/P(F)= 0.70 * 0.68/0.623 = 0.764 (76.4%)
where
P(S∩F)=probability that the study was sponsored by food industry and the product was found favorable
P(S/F)=probability that the study was sponsored by food industry given that the product was found favorable
c) for U= the product was found unfavourable , doing the same procedure as in a)
P(U)= 0.70 * 0.10 + 0.30 * 0.35 = 0.175
and the corresponding conditional probability is
P(N/U)=P(N∩U)/P(U)= 0.30 * 0.35 / 0.175 = 0.6 (60%)
where N represents the event = the study had no industry funding
10.5 - 8.7 = 1.8
There is 1.8 feet of pipe left
-(1.5, -3)
think of the y axis intercepting the x axis at coordinates (0,0)
-they then said their point is 1.5 left of the y axis
-now all you have to do is to find another x value of that same distance (x,y)
