The probability of selecting exactly one ace is its likelihood
The probability that a five-card poker hand contains exactly one ace is 29.95%
<h3>How to determine the probability?</h3>
There are 4 aces in a standard deck of 52 cards.
The probability of selecting an ace would be:
p = 4/52
Also, there are 48 non-ace cards in the standard deck
So, the probability of selecting a non-ace after an ace has been selected is:
p = 48/51
The probability of selecting a non-ace up to the fifth selection are:
- After two cards have been selected is: 47/50.
- After three cards have been selected is: 46/49.
- After four cards have been selected is: 45/48.
The required probability is then calculated as:
P(1 Ace) = n * (4/52) * (48/51) * (47/50) * (46/49) * (45/48)
Where n is the number of cards i.e. 5
So, we have:
P(1 Ace) = 5 * (4/52) * (48/51) * (47/50) * (46/49) * (45/48)
Evaluate
P(1 Ace) = 0.2995
Express as percentage
P(1 Ace) = 29.95%
Hence, the probability that a five-card poker hand contains exactly one ace is 29.95%
Read more about probability at:
brainly.com/question/25870256
Answer with Step-by-step explanation:
We are given that u and v are functions of x and both are differentiable at x=0

a.We have to find the values of 

Using this formula
Then , we get
![[\frac{d(uv)}{dx}]_{x=0}=u'(0)v(0)+u(0)v'(0)=7(2)+4(1)=14+4=18](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%28uv%29%7D%7Bdx%7D%5D_%7Bx%3D0%7D%3Du%27%280%29v%280%29%2Bu%280%29v%27%280%29%3D7%282%29%2B4%281%29%3D14%2B4%3D18)
![[\frac{d(uv)}{dx}]_{x=0}=18](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%28uv%29%7D%7Bdx%7D%5D_%7Bx%3D0%7D%3D18)
b.
![[\frac{d(u/v)}{dx}]_{x=0}=\frac{u'(0)v(0)-u(0)v'(0)}{v^2(0)}=\frac{7(2)-4(1)}{2^2}=\frac{14-4}{4}=\frac{10}{4}=\frac{5}{2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%28u%2Fv%29%7D%7Bdx%7D%5D_%7Bx%3D0%7D%3D%5Cfrac%7Bu%27%280%29v%280%29-u%280%29v%27%280%29%7D%7Bv%5E2%280%29%7D%3D%5Cfrac%7B7%282%29-4%281%29%7D%7B2%5E2%7D%3D%5Cfrac%7B14-4%7D%7B4%7D%3D%5Cfrac%7B10%7D%7B4%7D%3D%5Cfrac%7B5%7D%7B2%7D)
![[\frac{d(u/v)}{dx}]_{x=0}=\frac{5}{2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%28u%2Fv%29%7D%7Bdx%7D%5D_%7Bx%3D0%7D%3D%5Cfrac%7B5%7D%7B2%7D)
c.
![[\frac{d(v/u)}{dx}]_{x=0}=\frac{v'(0)u(0)-v(0)u'(0)}{u^2(0)}=\frac{1(4)-7(2)}{4^2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%28v%2Fu%29%7D%7Bdx%7D%5D_%7Bx%3D0%7D%3D%5Cfrac%7Bv%27%280%29u%280%29-v%280%29u%27%280%29%7D%7Bu%5E2%280%29%7D%3D%5Cfrac%7B1%284%29-7%282%29%7D%7B4%5E2%7D)
![[\frac{d(v/u)}{dx}]_{x=0}=\frac{-10}{16}=\frac{-5}{8}](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%28v%2Fu%29%7D%7Bdx%7D%5D_%7Bx%3D0%7D%3D%5Cfrac%7B-10%7D%7B16%7D%3D%5Cfrac%7B-5%7D%7B8%7D)
d.
![[\frac{d(-6v-9u)}{dx}]_{x=0}=-6v'(0)-9u'(0)=-6(1)-9(7)=-6-63=-69](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%28-6v-9u%29%7D%7Bdx%7D%5D_%7Bx%3D0%7D%3D-6v%27%280%29-9u%27%280%29%3D-6%281%29-9%287%29%3D-6-63%3D-69)
![[\frac{d(-6v-9u)}{dx}]_{x=0}=-69](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%28-6v-9u%29%7D%7Bdx%7D%5D_%7Bx%3D0%7D%3D-69)
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Answer:
The probability of failure of both the bulbs is 0.4323.
Step-by-step explanation:
For an exponential distribution the distribution is given by

The value of λ is related to the mean μ as λ=1/μ,
Let us denote the 2 bulbs by X and Y thus the probability distribution of the 2 bulbs is as under

Similarly for the bulb Y the distribution function is given by
Thus the probability for both the bulbs to fail within 1500 hours is
