Answer:
The required probability is
or
.
Step-by-step explanation:
Consider the provided information.
A committee of 9 members is voting on a proposal. Each member casts a yea or nay vote. On a random voting basis,
The probability of yea or nay vote is equal, = ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
So, we can say that ![p=q=\frac{1}{2}](https://tex.z-dn.net/?f=p%3Dq%3D%5Cfrac%7B1%7D%7B2%7D)
Use the formula: ![P(x)=\binom{n}{x}p^xq^{n-x}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cbinom%7Bn%7D%7Bx%7Dp%5Exq%5E%7Bn-x%7D)
Where n is the total number of trials, x is the number of successes, p is the probability of getting a success and q is the probability of failure.
We want proposal wins by a vote of 7 to 2, that means the value of x is 7.
Substitute the respective values in the above formula.
![P(x)=\binom{9}{7}(\frac{1}{2})^7(\frac{1}{2})^{9-7}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cbinom%7B9%7D%7B7%7D%28%5Cfrac%7B1%7D%7B2%7D%29%5E7%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B9-7%7D)
![P(x)=\frac{9!}{7!2!}(\frac{1}{2})^7(\frac{1}{2})^{2}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7B9%21%7D%7B7%212%21%7D%28%5Cfrac%7B1%7D%7B2%7D%29%5E7%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B2%7D)
![P(x)=\frac{8\times9}{2}\times(\frac{1}{2})^9](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7B8%5Ctimes9%7D%7B2%7D%5Ctimes%28%5Cfrac%7B1%7D%7B2%7D%29%5E9)
![P(x)=\frac{4\times9}{2^9}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7B4%5Ctimes9%7D%7B2%5E9%7D)
![P(x)=\frac{9}{2^7}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7B9%7D%7B2%5E7%7D)
or ![P(x)=0.0703125](https://tex.z-dn.net/?f=P%28x%29%3D0.0703125)
Hence, the required probability is
or
.
Answer:
2 hours
Step-by-step explanation:
first tap= 3hours
therefore work done is = 1/3
second tap= 6hours
therefore work done = 1/6
so, 1/3+ 1/6
=2/6 + 1/6
=3/6
=1/2
=2/1 hours
=2 hours
Answer:
n=6
m=4.5
I don't know about the angle though
Step-by-step explanation:
sf = 1.5
1.5x4=6
1.5x3= 4.5
Answer:
29.791
Step-by-step explanation:
3.1^3 = 29.791
set as brainliest
Answer:
4^7
Step-by-step explanation:
please mark me as brainlest