At Least 2 In Probability
"At least" and "at virtually," and mean, variance, and standard deviation
"At least" and "at nigh" probability
Nosotros can do more just calculate the probability of pulling exactly ???three??? red marbles in ???5??? total pulls.
For whatever binomial random variable, we can also calculate something like the probability of pulling at least ???3??? cherry marbles, or the probability of pulling no more than ???3??? marbles.
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What we want to know is that the probability of pulling at least ???iii??? red marbles is the probability that we pull ???3???, or ???iv???, or ???five??? red marbles, which is only the probability of each of these, all added together.
???P(\text{at to the lowest degree }3\text{ reds in }5\text{ pulls})=P(three\text{ reds})+P(4\text{ reds})+P(5\text{ reds})???
???P(\text{at to the lowest degree }3\text{ reds in }5\text{ pulls})=\binom{5}{iii}\left(\frac13\right)^iii 0.67^two???
???+\binom{5}{4}\left(\frac13\right)^4 \left(\frac23\right)^ane+\binom{5}{5}\left(\frac13\right)^five \left(\frac23\correct)^0???
???P(\text{at least }3\text{ reds in }v\text{ pulls})=(10)\left(\frac13\correct)^3\left(\frac23\right)^2???
???+(five)\left(\frac13\right)^four\left(\frac23\correct)^ane+(one)\left(\frac13\right)^v\left(\frac23\correct)^0???
???P(\text{at to the lowest degree }iii\text{ reds in }5\text{ pulls})\approx0.1646+0.0412+0.0041???
???P(\text{at least }three\text{ reds in }5\text{ pulls})\approx0.2099???
???P(\text{at least }3\text{ reds in }5\text{ pulls})\approx21\%???
In the same way, the probability of pulling at most ???3??? red marbles would be the probability of pulling ???one???, or ???ii???, or ???three??? red marbles, all added together.
If we're calculating the probability of at to the lowest degree one success or at least one failure, nosotros can use these formulas:
???P(\text{at to the lowest degree one success})=1-P(\text{all failures})???
???P(\text{at least 1 failure})=1-P(\text{all successes})???
This is because all probability distribution functions must add up to ???1???.
How to calculate the probability of "at least" and "at most" events
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Probability of getting at least 1 "heads" on 5 coin flips
Example
Detect the probability that nosotros get at least ???1??? heads on ???5??? coin flips.
We tin can actually simplify this problem a lot by realizing that every single prepare of ???v??? coin flips will have at least one heads, unless every one of the ???5??? flips is tails: ???TTTTT???. The probability of getting ???v??? tails in a row is
???P(TTTTT)=\left(\frac12\right)\left(\frac12\correct)\left(\frac12\right)\left(\frac12\correct)\left(\frac12\right)=\frac{1}{32}???
The probability of getting "at least i heads" is the same as the probability of non getting "all tails." Therefore, since full probability is always equal to ???1???, we can say that the probability of at least one heads is
???P(\text{at least 1 heads})=1-\frac{1}{32}=\frac{31}{32}???
Let'due south do some other example where nosotros detect an "at most" probability for a binomial random variable.
The probability of getting "at least i heads" is the same every bit the probability of not getting "all tails."
Example
Let ???Ten??? exist a binomial random variable with ???n=ten??? and ???p=0.thirty???. Discover ???P(10\le 5)???.
The variable ???X??? follows a binomial distribution, merely instead of finding the probability of exactly ???k??? successes in ???n??? trials, we're asked to discover the probability of ???1000??? or fewer successes in ???n??? trials. Specifically, observe the chance of ???5??? or fewer successes in ???10??? trials, where the probability of success on whatsoever ane trial is ???p=0.30???.
Find the probability of ???0??? successes, ???1??? success, ???2??? successes, etc., up to ???5??? successes, and then find the sum of those probabilities.
???P(X\le v)=P(Ten=0)+P(X=1)+P(X=two)+P(10=3)+P(X=four)+P(X=v)???
To find the probability for each value of ???k???, we use the binomial probability formula.
???P(k\text{ successes in }n\text{ trials})=\binom{n}{thou}p^1000(1-p)^{north-g}???
Then the probability ???P(X\le five)??? is
???P(X\le 5)=\binom{10}{0}(0.30)^0(1-0.30)^{10}+\binom{x}{1}(0.30)^ane(1-0.thirty)^{9}???
???+\binom{10}{2}(0.30)^{2}(1-0.thirty)^8+\binom{ten}{three}(0.30)^{three}(1-0.30)^vii???
???+\binom{10}{4}(0.thirty)^{iv}(1-0.30)^half-dozen+\binom{ten}{5}(0.30)^{5}(1-0.thirty)^five???
???P(X\le five)=0.9527???
Mean, variance, and standard deviation
The hateful of a binomial random variable ???X??? tin can be expressed as ???\mu_X???. The mean is too called the expected value, and that's indicated equally ???E(X)???. Either style, the mean is given by
???\mu_X=E(X)=np???
where ???n??? is the fixed number of contained trials, and ???p??? is the probability of a success. The variance of a binomial random variable ???10??? is given by
???\sigma_X^2=np(1-p)???
Standard deviation is the square root of the variance and is therefore given past
???\sqrt{\sigma_X^2}=\sqrt{np(1-p)}???
???\sigma_X=\sqrt{np(ane-p)}???
If nosotros continue with our instance of the number of heads we get on ???five??? money flips, we can say that the number of trials ???n??? is ???five???, and the probability of success (getting heads) is ???p=0.5???. Therefore, the mean is
???\mu_X=np???
???\mu_X=v(0.5)???
???\mu_X=ii.5???
The variance is
???\sigma_X^two=np(one-p)???
???\sigma_X^2=v(0.5)(i-0.v)???
???\sigma_X^2=ii.5(1-0.5)???
???\sigma_X^2=2.five(0.5)???
???\sigma_X^2=1.25???
And the standard deviation is
???\sigma_X=\sqrt{np(1-p)}???
???\sigma_X=\sqrt{1.25}???
???\sigma_X\approx1.12???
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At Least 2 In Probability,
Source: https://www.kristakingmath.com/blog/at-least-and-at-most-probability
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