Answer:
No solution
Step-by-step explanation:
Answer:
the answer is c = (ad)/b
Step-by-step explanation:
See the picture
Answer: a) 0.2222, b) 0.3292, c) 0.1111
Step-by-step explanation:
Since we have given that
Let the probability of getting head be p.
Since, its head is twice as likely to occur as its tail.
![p+\dfrac{p}{2}=1\\\\\dfrac{3p}{2}=1\\\\p=\dfrac{2}{3}](https://tex.z-dn.net/?f=p%2B%5Cdfrac%7Bp%7D%7B2%7D%3D1%5C%5C%5C%5C%5Cdfrac%7B3p%7D%7B2%7D%3D1%5C%5C%5C%5Cp%3D%5Cdfrac%7B2%7D%7B3%7D)
a)If the coin is flipped 3 times, what is the probability of getting exactly 1 head?
So, here, n = 3
![p=\dfrac{2}{3}](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7B2%7D%7B3%7D)
![q=\dfrac{1}{3}](https://tex.z-dn.net/?f=q%3D%5Cdfrac%7B1%7D%7B3%7D)
Now,
![P(X=1)=^3C_1(\dfrac{2}{3})^1(\dfrac{1}{3})^2=0.2222](https://tex.z-dn.net/?f=P%28X%3D1%29%3D%5E3C_1%28%5Cdfrac%7B2%7D%7B3%7D%29%5E1%28%5Cdfrac%7B1%7D%7B3%7D%29%5E2%3D0.2222)
b)If the coin is flipped 5 times, what is the probability of getting exactly 2 tails?
2 tails means 3 heads.
So, it becomes,
![P(X=3)=^5C_3(\dfrac{2}{3})^3(\dfrac{1}{3})^2=0.3292](https://tex.z-dn.net/?f=P%28X%3D3%29%3D%5E5C_3%28%5Cdfrac%7B2%7D%7B3%7D%29%5E3%28%5Cdfrac%7B1%7D%7B3%7D%29%5E2%3D0.3292)
c)If the coin is flipped 4 times, what is the probability of getting at least 3 tails?
![P(X\leq 1)=\sum _{x=0}^1^4C_x(\dfrac{2}{3}^x(\dfrac{1}{3})^{4-x}=0.1111](https://tex.z-dn.net/?f=P%28X%5Cleq%201%29%3D%5Csum%20_%7Bx%3D0%7D%5E1%5E4C_x%28%5Cdfrac%7B2%7D%7B3%7D%5Ex%28%5Cdfrac%7B1%7D%7B3%7D%29%5E%7B4-x%7D%3D0.1111)
Hence, a) 0.2222, b) 0.3292, c) 0.1111
2+ 0.0+ 0.05+ 0.009okay so you just have to add the numbers. so you add 2+0.0+0.5+0.009. then you will get you answer. i hope this helps!!!!
Using the formula for the future value of an annuity: FV = P x (1 + rate)^time - 1 / rate)
1st account:
15,000 x (1 + 0.05)^10 - 1 /0.05) = $188,668.39
2nd account:
15,000 x (1 + 0.10)^10 - 1 /0.10) = $239,061.37