Answer:
This experiment is like throwing 3 balls independently at random into 3 boxes labeled 1, 2, 3.
The probability that the number 105 can be formed, with rearrengement if necessary, is 0.006
Step-by-step explanation:
The probability that a 105 (in that order) happens is equal to the probability of any other arrangement of 105 (in fact, every number is equally likely to be selected without rearrangement).
The total amount of rearrengements for 105 is 3! = 6 (the first digit can be anyone from 1, 0 or 5, for the second one we have only 2 possibilities and from the third one only one). The probability for a specific rearrengement to be picked is (1/10)³ = 1/1000, because we only have 1 favourable case over 10 for each of the 3 digits, and we need the 3 to be favourable, thus we need to power 1/10 by 3.
Therefore, the probability that a rearrengement of 105 is obtained is 6* 1/1000 = 3/500 = 0.006.
Answer:
167/346 or 0.483
Step-by-step explanation:
From the question given above, the following data were obtained:
Number of Tails (T) = 167
Number of Heads (H) = 179
Probability of tail, P(T) =?
Next, we shall determine total outcome. This can be obtained as follow:
Number of Tails (T) = 167
Number of Heads (H) = 179
Total outcome (S) =?
S = T + H
S = 167 + 179
Total outcome (S) = 346
Finally, we shall determine the probability of tails. This can be obtained as follow:
Number of Tails (T) = 167
Total outcome (S) = 346
Probability of tail, P(T) =?
P(T) = T / S
P(T) = 167 / 346
P(T) = 0.483
Thus, the probability of tails is 167/346 or 0.483
Answer:
1. The probability that the student will get exactly 6 correct answers is
.
2. The probability that the student will get more than 6 correct answers is
.
Step-by-step explanation:
From the given information it is clear that
The total number of equations (n) = 10
The probability of selecting the correct answer (p)= ![\frac{1}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D)
The probability of selecting the incorrect answer (q)= ![1-p=1-\frac{1}{3}=\frac{2}{3}](https://tex.z-dn.net/?f=1-p%3D1-%5Cfrac%7B1%7D%7B3%7D%3D%5Cfrac%7B2%7D%7B3%7D)
According to the binomial distribution, the probability of selecting r items from n items is
![P=^nC_rp^rq^{n-r}](https://tex.z-dn.net/?f=P%3D%5EnC_rp%5Erq%5E%7Bn-r%7D)
where, p is probability of success and q is the probability of failure.
The probability that the student will get exactly 6 correct answers is
![P(r=6)=^{10}C_6(\frac{1}{3})^6(\frac{2}{3})^{10-6}](https://tex.z-dn.net/?f=P%28r%3D6%29%3D%5E%7B10%7DC_6%28%5Cfrac%7B1%7D%7B3%7D%29%5E6%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B10-6%7D)
![P(r=6)=210(\frac{1}{3})^6(\frac{2}{3})^{4}=\frac{1120}{19683}](https://tex.z-dn.net/?f=P%28r%3D6%29%3D210%28%5Cfrac%7B1%7D%7B3%7D%29%5E6%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B4%7D%3D%5Cfrac%7B1120%7D%7B19683%7D)
Therefore the probability that the student will get exactly 6 correct answers is
.
The probability that the student will get more than 6 correct answers is
![P(r>6)=^{10}C_7(\frac{1}{3})^7(\frac{2}{3})^{10-7}+^{10}C_8(\frac{1}{3})^8(\frac{2}{3})^{10-8}+^{10}C_9(\frac{1}{3})^9(\frac{2}{3})^{10-9}+^{10}C_{10}(\frac{1}{3})^{10}(\frac{2}{3})^{10-10}](https://tex.z-dn.net/?f=P%28r%3E6%29%3D%5E%7B10%7DC_7%28%5Cfrac%7B1%7D%7B3%7D%29%5E7%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B10-7%7D%2B%5E%7B10%7DC_8%28%5Cfrac%7B1%7D%7B3%7D%29%5E8%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B10-8%7D%2B%5E%7B10%7DC_9%28%5Cfrac%7B1%7D%7B3%7D%29%5E9%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B10-9%7D%2B%5E%7B10%7DC_%7B10%7D%28%5Cfrac%7B1%7D%7B3%7D%29%5E%7B10%7D%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B10-10%7D)
![P(r>6)=^{10}C_7(\frac{1}{3})^7(\frac{2}{3})^{3}+^{10}C_8(\frac{1}{3})^8(\frac{2}{3})^{2}+^{10}C_9(\frac{1}{3})^9(\frac{2}{3})^{1}+^{10}C_{10}(\frac{1}{3})^{10}(\frac{2}{3})^{0}](https://tex.z-dn.net/?f=P%28r%3E6%29%3D%5E%7B10%7DC_7%28%5Cfrac%7B1%7D%7B3%7D%29%5E7%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B3%7D%2B%5E%7B10%7DC_8%28%5Cfrac%7B1%7D%7B3%7D%29%5E8%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B2%7D%2B%5E%7B10%7DC_9%28%5Cfrac%7B1%7D%7B3%7D%29%5E9%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B1%7D%2B%5E%7B10%7DC_%7B10%7D%28%5Cfrac%7B1%7D%7B3%7D%29%5E%7B10%7D%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B0%7D)
![P(r>6)=120\times \frac{8}{59049}+45\times \frac{4}{59049}+10\times \frac{2}{59049}+1\times \frac{1}{59049}=\frac{43}{2187}](https://tex.z-dn.net/?f=P%28r%3E6%29%3D120%5Ctimes%20%5Cfrac%7B8%7D%7B59049%7D%2B45%5Ctimes%20%5Cfrac%7B4%7D%7B59049%7D%2B10%5Ctimes%20%5Cfrac%7B2%7D%7B59049%7D%2B1%5Ctimes%20%5Cfrac%7B1%7D%7B59049%7D%3D%5Cfrac%7B43%7D%7B2187%7D)
Therefore the probability that the student will get more than 6 correct answers is
.
Answer:
Three partners share for business. Each partner’s receive Max-$ 93,750 and Sherry-$ 100,000 and Duane-$ 56,250
<u>Solution:</u>
Given that three partners share the business
Max owns 3/8, Sherry owns 2/5, and Duane owns the rest
Profit of the year is 250,000
We need to find how much is the share of each partner
Share owned by Max ![=\frac{3}{8}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B3%7D%7B8%7D)
Share owned by Sharry = ![\frac{2}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B5%7D)
Share owned by Duane ![=1-\left(\frac{3}{8}+\frac{2}{5}\right)=\frac{9}{40}](https://tex.z-dn.net/?f=%3D1-%5Cleft%28%5Cfrac%7B3%7D%7B8%7D%2B%5Cfrac%7B2%7D%7B5%7D%5Cright%29%3D%5Cfrac%7B9%7D%7B40%7D)
Total profit earned = $250,000
Thus, Max receives =
= $93,750
Sharry receives =
= $100,000
Duane receives =
= $56,250
Hence the share of Max , Sharry and Duane is calculated.
Answer:
If a person is randomly selected from this group, the probability that they have both high blood pressure and high cholesterol is P=0.25.
Step-by-step explanation:
We can calculate the number of people from the sample that has both high blood pressure (HBP) and high cholesterol (HC) using this identity:
![N(\text{HBP or HC})=N(\text{HBP})+N(\text{HC})-N(\text{HBP and HC})\\\\\\ N(\text{HBP and HC})=N(\text{HBP})+N(\text{HC})-N(\text{HBP or HC})\\\\\\ N(\text{HBP and HC})=15+25-30=10](https://tex.z-dn.net/?f=N%28%5Ctext%7BHBP%20or%20HC%7D%29%3DN%28%5Ctext%7BHBP%7D%29%2BN%28%5Ctext%7BHC%7D%29-N%28%5Ctext%7BHBP%20and%20HC%7D%29%5C%5C%5C%5C%5C%5C%20N%28%5Ctext%7BHBP%20and%20HC%7D%29%3DN%28%5Ctext%7BHBP%7D%29%2BN%28%5Ctext%7BHC%7D%29-N%28%5Ctext%7BHBP%20or%20HC%7D%29%5C%5C%5C%5C%5C%5C%20N%28%5Ctext%7BHBP%20and%20HC%7D%29%3D15%2B25-30%3D10)
We can calculate the probability that a random person has both high blood pressure and high cholesterol as:
![P(\text{HBP and HC})=\dfrac{10}{40}=0.25](https://tex.z-dn.net/?f=P%28%5Ctext%7BHBP%20and%20HC%7D%29%3D%5Cdfrac%7B10%7D%7B40%7D%3D0.25)