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## There are two boxes: one contains cards numbered from 1 to 11 and other contains cards numbered from 1-5. A box is selected at random and one card is drawn. If it is even, then another card is drawn from either of the two boxes. Find P (both are even)?

Parzival S.
Featured 6 months ago

#(15,557)/(96,800)~=16.07%#

#### Explanation:

Ok... let's walk through this one slowly...

The first thing we have is 2 boxes, which I'll call Box A (which holds 11 cards) and Box B (which holds 5 cards).

$P \left(\text{Box A")=P("Box B}\right) = \frac{1}{2}$

Draw 1, Box A, chance of drawing an even card

Let's say we draw from Box A. We have cards numbered 1-11, and of the 11 cards, 5 are even. And so we can say that if we draw from Box A, the probability of drawing an even card is $\frac{5}{11}$

Draw 1, Box B, chance of drawing an even card

But what if instead we draw from Box B. We have cards numbered 1-5, and of the 5 cards, 2 are even. And so we can say that if we draw from Box B, the probability of drawing an even card is $\frac{2}{5}$

Draw 1, probability of drawing an even card

We can now put the above together to calculate the probability of drawing an even card:

$P \left(\text{drawing an even card}\right) = \frac{1}{2} \times \frac{5}{11} + \frac{1}{2} \times \frac{2}{5} = \frac{5}{22} + \frac{1}{5} \implies$

$\implies \frac{5}{22} \left(\frac{5}{5}\right) + \frac{1}{5} \left(\frac{22}{22}\right) = \frac{25}{110} + \frac{22}{110} = \frac{47}{110}$

We can now move on to the second draw.

Draw 2, Box A (where draw 1 was also Box A)

In this possibility, we drew from Box A in draw 1 and are doing so again in draw 2. There are now 10 cards and 4 of them are even, and so we get $\frac{4}{10} = \frac{2}{5}$

Draw 2, Box A (where draw 1 was Box B)

In this possibility, since we drew from Box B before, Box A still has 11 cards and 5 evens, and so that's $\frac{5}{11}$

Draw 2, Box B (where draw 1 was also Box B)

In this possibility, we drew from Box B in draw 1 and are doing so again in draw 2. There are now 4 cards and 1 of them is even, and so we get $\frac{1}{4}$

Draw 2, Box B (where draw 1 was Box A)

In this possibility, since we drew from Box A before, Box B still has 5 cards and 2 evens, and so that's $\frac{2}{5}$

Draw 2, probability of drawing an even card

All four of the above possibilities is equally likely, and so we'll multiply each of the above probabilities by $\frac{1}{4}$ and then sum them up:

$P \left(\text{drawing an even card on second draw}\right) = \frac{1}{4} \times \frac{2}{5} + \frac{1}{4} \times \frac{5}{11} + \frac{1}{4} \times \frac{2}{5} + \frac{1}{4} \times \frac{1}{4} = \frac{1}{10} + \frac{5}{44} + \frac{1}{10} + \frac{1}{16} \implies$

$\frac{1}{10} \left(\frac{88}{88}\right) + \frac{5}{44} \left(\frac{20}{20}\right) + \frac{1}{10} \left(\frac{88}{88}\right) + \frac{1}{16} \left(\frac{55}{55}\right)$

$\frac{88}{880} + \frac{100}{880} + \frac{88}{880} + \frac{55}{880} = \frac{331}{880}$

And now we can finalize this calculation by multiplying in the probability of getting an even card on the first draw:

#P("drawing two even cards")=47/110xx331/880=(15,557)/(96,800)~=16.07%#

## The probability of a winning certain game is 3/7. if he plays the game thrice, find the probability that a) he wins only the first game, b) he wins exactly once, and c) he wins in two out of three games?

Will
Featured 4 months ago

a) #~~42.86%#
b) #~~41.98%#
c) #~~39.36%#

#### Explanation:

Probability of winning the first game means he must win the first and the outcome of winning or losing beyond that is not meaningful so let's think about possible outcomes

• $w 1 =$ win or lose
• $w 2 =$ win or lose
• $w 3 =$ win or lose

so in the case that w1=win then w2 and w3 can be a win or lose. This means that the total outcomes is

$p \left(w 1 = \text{win" and w2="win" and w3="win}\right) = \frac{3}{7} \cdot \frac{3}{7} \cdot \frac{3}{7}$

let's just say

$P \left(w 1 = w , w 2 = w , w 3 = w\right) = {\left(\setminus \frac{3}{7}\right)}^{3}$

another outcome is
$P \left(w 1 = w , w 2 = w , w 3 = l\right) = \setminus \frac{3}{7} \cdot \setminus \frac{3}{7} \cdot \setminus \frac{4}{7}$

continuing this line of thinking the other two outcomes where w1 wins is thus

$P \left(w 1 = w , w 2 = l , w 3 = l\right) = \setminus \frac{3}{7} \cdot \setminus \frac{4}{7} \cdot \setminus \frac{4}{7}$
and
$P \left(w 1 = w , w 2 = l , w 3 = w\right) = \setminus \frac{3}{7} \cdot \setminus \frac{4}{7} \cdot \setminus \frac{3}{7}$

now if we want to determine winning the first game we can consider the 4 outcomes or $P \left(S 1 \setminus \mathmr{and} S 2 \setminus \mathmr{and} S 3 \mathmr{and} S 4\right)$, where $S$ stands for scenario. You will notice there are 4 scenarios so. since these are all mutually exclusive we should be able to add the probabilities thus

a) =#\frac{27}{343} + \frac{36}{343} + \frac{36}{343} +\frac{48}{343} ~~ 42.86%#
you notice this is the same result for probability of winning a game. This makes sense because if we were to construct a tree you would see that winning the first game is always $\frac{3}{7}$ regardless of the path after it, but its a good exercise to see this.

the probability of winning the game exactly once is the same as
$P \left(w 1 = w , w 2 = l , w 3 = l\right) \mathmr{and} P \left(w 1 = l , w 2 = w , w 3 = l\right) \mathmr{and} P \left(w 1 = l , w 2 = l , w 3 = w\right)$.

b = #3* \frac{48}{343} ~~41.98% #

the probability of winning 2 out of 3 games the same as
$P \left(w 1 = w , w 2 = w , w 3 = l\right) \mathmr{and} P \left(w 1 = l , w 2 = w , w 3 = w\right) \mathmr{and} P \left(w = w , w 2 = l , w 3 = w\right) \mathmr{and} P \left(w 1 = w , w 2 = w , w 3 = w\right)$.

c = #3* \frac{36}{343} +\frac{27}{343} ~~39.36% #

it might not seem intuitive that we include the scenario where we win in all 3 games but this is because this is a valid outcome for winning 2 out of 3 times and the question is not exactly 2 out of 3 times

## Let π denote the proportion of grocery store customers that use the store's club card. For a large sample z test of Hₒ: π = 0.5 versus Hₐ: π > 0.5, how do you find the P-value associated with the test statistic .93?

RomanSage
Featured 4 months ago

$p - v a l u e = 0.176$

#### Explanation:

(1) Hypotheses: ${H}_{0} : \pi = 0.5$ and ${H}_{1} : \pi > 0.5$

(2) Null Distribution: One-tailed standardized normal distribution

(3) Test statistics: $0.93$ (that is given to us)

(4) Decision Rule: Accept ${H}_{0}$ if p-value is greater than level of significance

For this question no level of significance is given; we could assume it is #95%#. But it is mandatory for the question to let know the level of significance or else no proper conclusion will be
applicable.

(5) Evidence: p-value is given by $P \left(x > 0.93\right) = 0.176$ with $\sigma = 1$ $\mu = 0$ (as of standardized normal distirbution)

(6) Decision: Do not reject ${H}_{0}$ as p-value is greater than $0.05$

(7) Conclusion: At 95% level of significance there is insufficient evidence to favor ${H}_{1}$ as the p-value is greater than 0.05 (one-tailed distribution with 95% level of significance). Therefore, the hypothesis that $\pi = 0.05$ is correct at this level of significance!

Assumptions

• The distribution is standardized normal
• 95 % level of significance is used

## Experimental/observational studies & relative frequencies?

Geoff K.
Featured 2 months ago

See explanation.

#### Explanation:

An experimental study is one where a test of some sort needs to be given in order to produce a measurement. Neither the subject nor the experimenter could possibly know the exact outcome before it is measured. The outcome for each subject may also change if the test is done again. Good examples are when the effects of a pill are tested, or a speed test is given.

Observational studies are those where we simply observe a trait, opinion, or measurement for a subject; the subject may know the outcome before the observation is made, and the outcome for each subject is not expected to change through repeated observations. Good examples would be a questionnaire, opinion poll, or height/weight measurements.

9) is an example of an experimental study, while 10) is an example of an observational study.

A frequency distribution counts how many measurements fall within each of the categories the experimenter is interested in. It can be written as a table, but is most often expressed as a bar graph.

For 11), the frequency distribution could be written in a table, like this:

$7$ Red
$3$ Yellow
$1$ Green
$4$ Blue
$5$ Purple

But since we tend to infer more useful information from a graph than a table, the bar graph for this data would look like this:

The relative frequencies are the ratios of the number in each category to the total number of measurements. For example, since there are 7 preschoolers whose favourite colour is red, and the total number of preschoolers surveyed is 20 (7+3+1+4+5), the relative frequency for red is $\frac{7}{20}$, or 0.350.

All of the relative frequencies are:

$0.350$ Red
$0.150$ Yellow
$0.050$ Green
$0.200$ Blue
$0.250$ Purple

Notice how the sum of these numbers is $1$. This must be the case for all relative frequency tables, because we're listing the percentages of the class that named each colour as their favourite.

## A bag contain 8blue balls and 5 black balls .2successive draws of 3 balls are made without replacement. Find the probability that the first drawing will give 3 black balls and the second 3 blue balls ?

Featured 1 week ago

The probability is $= \frac{7}{429}$

#### Explanation:

First Draw

$8$ blue balls and $5$ black balls

The number of ways of getting $3$ balls from $13$ balls is

#=((13),(3))=(13!)/((13-3)!(3!))=(13*12*11)/(1*2*3)=286#

The number of ways of getting $3$ black balls from $5$ black balls is

#=((5),(3))=(5!)/((5-3)!(3!))=(5*4)/(2*1)=10#

The probability of getting $3$ black balls is

$= \frac{10}{286}$

Second Draw

$8$ blue balls and $2$ black balls

The number of ways of getting $3$ balls from $10$ balls is

#=((10),(3))=(10!)/((10-3)!(3!))=(10*9*8)/(1*2*3)=120#

The number of ways of getting $3$ blue balls from $8$ black balls is

#=((8),(3))=(8!)/((8-3)!(3!))=(8*7*6)/(3*2*1)=56#

The probability of getting $3$ black balls is

$= \frac{56}{120}$

Therefore,

The probability of getting $3$ black balls and $3$ blue balls is

$= \frac{10}{286} \cdot \frac{56}{120} = \frac{7}{429}$

## In a survey 21 people like product A,26 like B and like C.If14 people like A and B ,12 people like C and A,14 people B and C and 8 like all.Find how many like prdct C only.?

Parzival S.
Featured 6 days ago

21

#### Explanation:

For each product, there will be people who like it only (ex. A only), people who like it and one other (ex. A and B only) and those who like all three. Let's work through this.

I'll note that this question appears to be a more fleshed-out version of this one posted earlier in the week: https://socratic.org/questions/in-a-survey-it-was-found-that-21-people-liked-product-a-26-liked-product-b-and-3 where the total number of likes for C was 39. I'll use that figure in here.

For product A, 21 people in total like it. 14 people like A and B (and might like C), 12 like C and A (and might like B), and 8 like A, B and C. This means that we need to subtract the 8 (those who like all three) from the numbers listed for people liking two products. This will give us:

People who like A and B only $= 14 - 8 = 6$

People who like A and C only $= 12 - 8 = 4$

${\overbrace{\text{Likes A")^(21)-overbrace("Likes A, B only")^(6)-overbrace("Likes A, C only")^(4)-overbrace("Likes A, B, C ")^8=overbrace("Likes A only}}}^{\textcolor{red}{3}}$

For Product B, 26 people like it. 14 like A and B (and might like C), 14 people like B and C (and might like A), and 8 like all three. We can do the same sort of math for this that we did for A:

People who like A and B only $= 14 - 8 = 6$

People who like B and C only $= 14 - 8 = 6$

${\overbrace{\text{Likes B")^(26)-overbrace("Likes A, B only")^(6)-overbrace("Likes B, C only")^(6)-overbrace("Likes A, B, C ")^8=overbrace("Likes B only}}}^{\textcolor{red}{6}}$

We can now do C. 39 people like C. 4 people like C and A only. 6 people like B and C only. And 8 like A, B, and C. Therefore:

${\overbrace{\text{Likes C")^(39)-overbrace("Likes A, C only")^(4)-overbrace("Likes B, C only")^(6)-overbrace("Likes A, B, C ")^8=overbrace("Likes C only}}}^{\textcolor{red}{21}}$

~~~~~

I want to note that when I first started working with A that if we take the numbers given as "only", we'd end up with something that made no sense:

A total likes = 21

A, B likes = 14
A, C likes = 12

which puts us at 26 likes, which is more than the 21 given. And so I realized we needed to subtract out from the A, B likes, for instance, those people who also liked C.

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