You roll a standard number cube. Find P(number greater than 1).

1 answer:

6 0

Well if its a standard die then the numbers on the die would be 1,2,3,4,5,6. Since you want the probability that the number is GREATER than 1, you need to roll a 2,3,4,5, or 6. Which you have 5 total outcomes that would be successful out of 6. Makes the probability 5/6 or .833333

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According to the general equation for conditional probability, if P(A n B) = 3/10 and P(B)= 2/5 what is p(alb)

Natali5045456 [20]

Answer:

P(A given B) = 3/4

Step-by-step explanation:

As we know that it is conditional probability, where the probability of an event depends on the event that has certain probability of occurrence.

The formula for conditional probability is:

P(A given B) = P(A ∩B) / P(B)

Where a and B are events. The probability of event B is known and we also know probability of A∩B

So, putting the values in the formula:

$P(A\ given\ B) = \frac{P(A and B)}{P(B)}\\= \frac{\frac{3}{10} }{\frac{2}{5} } \\=\frac{3}{10} *\frac{5}{2}\\=\frac{3}{4}$

So, the probability of A given B is 3/4 ..

3 0

3 years ago

Suppose u1, u2, ..., un are independent random variables and for every i = 1, ..., n, ui has a uniform distribution over [0, 1].

sattari [20]

$Z=U_{(1)}=\min\{U_1,\ldots,U_n\}$

has CDF

$F_Z(z)=1-(1-F_{U_i}(z))^n$

where $F_{U_i}(u_i)$ is the CDF of $U_i$ . Since $U_i$ are iid. with the standard uniform distribution, we have

$F_{U_i}(u_i)=\begin{cases}0&\text{for }u_i$

and so

$F_Z(z)=1-(1-F_{U_i}(z))^n=\begin{cases}0&\text{for }z$

Differentiate the CDF with respect to $z$ to obtain the PDF:

$f_Z(z)=\dfrac{\mathrm dF_Z(z)}{\mathrm dz}=\begin{cases}n(1-z)^{n-1}&\text{for }0$

i.e. $Z$ has a Beta distribution $\beta(1,n)$ .

3 0

2 years ago

Brian borrowed a 20 foot extension ladder to use when he paints hes house. If he sets the base of the ladder 6 feet from the hou

iren [92.7K]

Answer:

It is 20.9feet

Step-by-step explanation: