Answer:
a) The probability that at least 5 ties are too tight is P=0.0432.
b) The probability that at most 12 ties are too tight is P=1.
Step-by-step explanation:
In this problem, we could represent the proabilities of this events with the Binomial distirbution, with parameter p=0.1 and sample size n=20.
a) We can express the probability that at least 5 ties are too tight as:
![P(x\geq5)=1-\sum\limits^4_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\geq5)=1-(0.1216+0.2702+0.2852+0.1901+0.0898)\\\\P(x\geq5)=1-0.9568=0.0432](https://tex.z-dn.net/?f=P%28x%5Cgeq5%29%3D1-%5Csum%5Climits%5E4_%7Bk%3D0%7D%20%7B%5Cfrac%7Bn%21%7D%7Bk%21%28n-k%29%21%7D%20p%5Ek%281-p%29%5E%7Bn-k%7D%7D%5C%5C%5C%5CP%28x%5Cgeq5%29%3D1-%280.1216%2B0.2702%2B0.2852%2B0.1901%2B0.0898%29%5C%5C%5C%5CP%28x%5Cgeq5%29%3D1-0.9568%3D0.0432)
The probability that at least 5 ties are too tight is P=0.0432.
a) We can express the probability that at most 12 ties are too tight as:
![P(x\leq 12)=\sum\limits^{12}_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\leq 12)=0.1216+0.2702+0.2852+0.1901+0.0898+0.0319+0.0089+0.0020+0.0004+0.0001+0.0000+0.0000+0.0000\\\\P(x\leq 12)=1](https://tex.z-dn.net/?f=P%28x%5Cleq%2012%29%3D%5Csum%5Climits%5E%7B12%7D_%7Bk%3D0%7D%20%7B%5Cfrac%7Bn%21%7D%7Bk%21%28n-k%29%21%7D%20p%5Ek%281-p%29%5E%7Bn-k%7D%7D%5C%5C%5C%5CP%28x%5Cleq%2012%29%3D0.1216%2B0.2702%2B0.2852%2B0.1901%2B0.0898%2B0.0319%2B0.0089%2B0.0020%2B0.0004%2B0.0001%2B0.0000%2B0.0000%2B0.0000%5C%5C%5C%5CP%28x%5Cleq%2012%29%3D1)
The probability that at most 12 ties are too tight is P=1.
I don’t get it but probably the scissors lol
One and only one line contains both A and B (assuming, that A and B do not overlap)
Answer:
a) 0.1091
b) 0.9994
c) 0.5886
Step-by-step explanation:
X = the number of fish out of 20 that die after 24 hours
x = 0, 1, 2, . . . , 20
X~ Binomial (n= 20, p =0.20)
P(14 survive) = P(X = 6)
=
=0.1091
Similarly we can find out
P(at least 10 survive) = P( X <= 10 ) = (Using technology) = 0.9994
P(at most 16 will survive) = P(X <= 16) = (Using technology) = 0.5886