Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
1 answer:
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:
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Step-by-step explanation:
The simplest method is "brute force". Calculate each term and add them up.
∑ = 3(1) + 3(2) + 3(3) + 3(4) + 3(5)
∑ = 3 + 6 + 9 + 12 + 15
∑ = 45
∑ = (2×1)² + (2×2)² + (2×3)² + (2×4)²
∑ = 4 + 16 + 36 + 64
∑ = 120
∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)
∑ = -4 + -2 + 0 + 2
∑ = -4
4. 1 + 1/4 + 1/16 + 1/64 + 1/256
This is a geometric sequence where the first term is 1 and the common ratio is 1/4. The nth term is:
a = 1 (1/4)ⁿ⁻¹
So the series is:
5. -5 + -1 + 3 + 7 + 11
This is an arithmetic sequence where the first term is -5 and the common difference is 4. The nth term is:
a = -5 + 4(n−1)
a = -5 + 4n − 4
a = 4n − 9
So the series is:
Answer:
Only A, E, and F are correct.
Step-by-step explanation:
The difference between two points is , thus the difference between the height of the Pelican and the height of the Heron is
, and between the Pelican and the trout it is
.
The distance between two points is just the absolute value of the difference between them. Between the Pelican and the Heron it is, and the distance between the Pelican and the Trout is
.
Therefore,
A is correct;
B is incorrect (difference is not positive);
C is incorrect (distance cannot be negative);
D is incorrect (difference is not positive);
E is correct;
F is correct;