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2 years ago

6

there are 5 red counters and 3 blue counters in bag A. there are 4 red counters in bag B. Tina takes at random a counter from ea

ch bag. what is the probability that Tina takes two blue counters

1 answer:

mars1129 [50]2 years ago

7 0

Answer:

The probability is 0

Step-by-step explanation:

From the first bag, the total number of counters is 5 + 3 = 8

The probability of taking a blue counter from the first bag is the number of blue counters divided by the total number of counters

Mathematically, we have this as;

3/8

For the second bag, all the counters are red with no blue counter

So the probability of taking a blue counter here is 0

Now, the probability that he takes a blue counter in A and a blue counter in B will

be;

3/8 * 0 = 0

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A coin is being flipped 400 times. What is the probability of it landing on heads at most 170 times, rounded to the nearest tent

lisabon 2012 [21]

Hello!

We have two probabilities we can use; we have 170/400, for our experiment, and 1/2, which is our theoretical probability.

To solve, we just multiply the two probabilities.

$\frac{170}{400}(1/2)$ =0.2125≈21.3

Therefore, we have about a 21.3% chance of this event occurring.

I hope this helps!

8 0

2 years ago

How many different 10 person committees can be selected from a pool of 23 people?

Salsk061 [2.6K]

We're going to be using combination since this question is asking how many different combinations of 10 people can be selected from a set of 23.

We would only use permutation if the order of the people in the committee mattered, which it seems it doesn't.

Formula for combination:

$C(n,r)=\dfrac{n!}{(n-r)!r!}$

Where $n$ represents the number of objects/people in the set and $r$ represents the number of objects/people being chosen from the set

There are 23 people in the set and 10 people being chosen from the set

$C(23,10)=\dfrac{23!}{(23-10)!10!}$

$=\dfrac{23!}{13!\times10!}$

Usually I would prefer solving such fractions by hand instead of a calculator, but factorials can result in large numbers and there is too much multiplication. Using a calculator, we get

$=1,144,066$

Thus, there are 1,144,066 different 10 person committees that can be selected from a pool of 23 people. Let me know if you need any clarifications, thanks!

~ Padoru

4 0

3 years ago

Read 2 more answers

A data set has a median of 12, an upper quartile of 15, a lower quartile of 10, a minimum of 4, and a maximum of 20. Which state

ololo11 [35]

Given:

A data set has a median of 12, an upper quartile of 15, a lower quartile of 10, a minimum of 4, and a maximum of 20.

To find:

The correct statement for the box plot.

Solution:

Lower quartile is 10 and upper quartile is 15, so the box will go from 10 to 15.

Median of the data set is 12, so a line dividing the box will be at 12.

Minimum value is 4 and lower quartile is 10, so the left whisker will go from 4 to 10.

Upper quartile is 15 and maximum value is 20, so the right whisker will go from 15 to 20.

Therefore, the correct option is B.

8 0

2 years ago

Read 2 more answers

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

3 years ago

..........................

Rainbow [258]

Option A

x+8=-3
x+8-8=-3-8
x=-11

Hope I didn't mess up for your sake!

7 0

3 years ago