I’m not sure but I need points so please mark this as answer :)
Answer:
(a) 0.28347
(b) 0.36909
(c) 0.0039
(d) 0.9806
Step-by-step explanation:
Given information:
n=12
p = 20% = 0.2
q = 1-p = 1-0.2 = 0.8
Binomial formula:

(a) Exactly two will be drunken drivers.



Therefore, the probability that exactly two will be drunken drivers is 0.28347.
(b)Three or four will be drunken drivers.


Using binomial we get



Therefore, the probability that three or four will be drunken drivers is 0.3691.
(c)
At least 7 will be drunken drivers.

![P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]](https://tex.z-dn.net/?f=P%28x%5Cleq%207%29%3D1-%5BP%28x%3D0%29%2BP%28x%3D1%29%2BP%28x%3D2%29%2BP%28x%3D3%29%2BP%28x%3D4%29%2BP%28x%3D5%29%2BP%28x%3D6%29%5D)
![P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]](https://tex.z-dn.net/?f=P%28x%5Cleq%207%29%3D1-%5B0.06872%2B0.20616%2B0.28347%2B0.23622%2B0.13288%2B0.05315%2B0.0155%5D)
![P(x\leq 7)=1-[0.9961]](https://tex.z-dn.net/?f=P%28x%5Cleq%207%29%3D1-%5B0.9961%5D)

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.
(d) At most 5 will be drunken drivers.



Therefore, the probability of at most 5 will be drunken drivers is 0.9806.
The answer to the question
Answer:
In general, IBM MQ object names can be up to 48 characters long
Yes it does because the line is straight.