Answer:
B. 0.03110
Step-by-step explanation:
Given
Probability of Hit = 60%
Required
Determine the probability that he misses at 6th throw
Represent Probability of Hit with P
![P = 60\%](https://tex.z-dn.net/?f=P%20%3D%2060%5C%25)
Convert to decimal
![P = 0,6](https://tex.z-dn.net/?f=P%20%3D%200%2C6)
Next; Determine the Probability of Miss (q)
Opposite probabilities add up to 1;
So,
![p + q = 1](https://tex.z-dn.net/?f=p%20%2B%20q%20%3D%201)
![q = 1 - p](https://tex.z-dn.net/?f=q%20%3D%201%20-%20p)
Substitute 0.6 for p
![q = 1 - 0.6](https://tex.z-dn.net/?f=q%20%3D%201%20-%200.6)
![q = 0.4](https://tex.z-dn.net/?f=q%20%3D%200.4)
Next,is to determine the required probability;
Since, he's expected to miss the 6th throw, the probability is:
![Probability = p^5 * q](https://tex.z-dn.net/?f=Probability%20%3D%20p%5E5%20%2A%20q)
![Probability = 0.6^5 * 0.4](https://tex.z-dn.net/?f=Probability%20%3D%200.6%5E5%20%2A%200.4)
![Probability = 0.031104](https://tex.z-dn.net/?f=Probability%20%3D%200.031104)
Hence;
<em>Option B answers the question</em>
Answer:
y
=
−
2
x
3
+
1
3
y
=
−
3
x
5
+
1
5
Step-by-step explanation:
Answer:
expand the equation
then make q the subject of the formal
Answer:
The probability of drawing an odd numbered ticket is 60%.
Step-by-step explanation:
Odd numbered tickets:
Probability of one is 1/5 plus half of 1/5.
![P(X = 1) = \frac{1}{5} + \frac{1}{10} = \frac{3}{10}](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7B1%7D%7B5%7D%20%2B%20%5Cfrac%7B1%7D%7B10%7D%20%3D%20%5Cfrac%7B3%7D%7B10%7D)
Probability of 3 is half of 1/5.
![P(X = 3) = \frac{1}{10}](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7B1%7D%7B10%7D)
Probability of 5 is 1/5. So
![P(X = 5) = \frac{1}{5}](https://tex.z-dn.net/?f=P%28X%20%3D%205%29%20%3D%20%5Cfrac%7B1%7D%7B5%7D)
Probability of drawing an odd numbered ticket:
![p = P(X = 1) + P(X = 3) + P(X = 5) = \frac{3}{10} + \frac{1}{10} + \frac{1}{5} = \frac{4}{10} + \frac{2}{10} = \frac{6}{10} = 0.6](https://tex.z-dn.net/?f=p%20%3D%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%205%29%20%3D%20%5Cfrac%7B3%7D%7B10%7D%20%2B%20%5Cfrac%7B1%7D%7B10%7D%20%2B%20%5Cfrac%7B1%7D%7B5%7D%20%3D%20%5Cfrac%7B4%7D%7B10%7D%20%2B%20%5Cfrac%7B2%7D%7B10%7D%20%3D%20%5Cfrac%7B6%7D%7B10%7D%20%3D%200.6)
0.6*100% = 60%
The probability of drawing an odd numbered ticket is 60%.
Answer:
Q=4
15q-16=8q+12
7q-16=12
7q=28
Q=4