Which equation can be used to solve for c? Triangle A B C is shown. Angle A C B is 90 degrees and angle A B C is 35 degrees. The
length of C B is 5 inches, the length of A C is b, and the length of B A is c. c = (5)cos(35o) c = StartFraction 5 Over cosine (35 degrees) EndFraction c = (5)sin(35o) c = StartFraction 5 Over sine (35 degrees) EndFraction
2 answers:
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Answer:
a) P=0
b) P=0.164
c) P=0.145
Step-by-step explanation:
We have 12 pieces, with 3 of each of the 4 flavors.
You draw the first 4 pieces.
a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.
b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.
The first probability is for the first draw, and has a value of 1, as any flavor will be ok.
The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.
The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.
The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.
Then, the probabilty of picking the 4 from different flavors is:
c) We can repeat the method for the previous probabilty.
The first draw has a probability of 1 because any flavor is ok.
In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.
For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.
If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.
For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.
For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.
We have then the probabilty as: