Question 195296
# 1


a)


*[Tex \LARGE P(\text{Bloom})=\frac{\text{Number of Tulips that Bloomed}}{\text{Number of Tulips Total}}] ... Start with the given probability



*[Tex \LARGE P(\text{Bloom})=\frac{62}{74}] ... Plug in the given values.



*[Tex \LARGE P(\text{Bloom})=\frac{31}{37}] ... Reduce.



*[Tex \LARGE P(\text{Bloom})=0.8378] ... Use a calculator (or long division) to find the approximate answer.



So the chances of a tulip blooming are 0.8378 which is 83.78% (multiply by 100 to get a percentage).


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b)



If we multiply the number planted by the probability that a flower will bloom, we will get an approximate number that will bloom. This is because the probability allows us to estimate a population.


So let x = number originally planted and y = number of blooming flowers wanted. 



So the equation is {{{y=P(bloom)*x}}}



{{{y=P(bloom)*x}}} Start with the given equation



{{{95=(0.8378)*x}}} Plug in {{{P(bloom)=0.8378}}} (the probability we found earlier) and {{{y=95}}} (the number of tulips wanted)



{{{95/0.8378=x}}} Divide both sides by 0.8378



{{{113.3922=x}}} Divide



{{{x=113.3922}}} Rearrange the equation.



{{{x=114}}} Round up to the nearest whole number (this guarantees that we pass our target).



So if the gardener plants 114 or more bulbs, then at least 95 bulbs should bloom.



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# 2


Notation note: The notation n(...) means "the number of ...". So n(assualts) is read as "the number of assualts" and n(violent crimes) is read as "the number of violent crimes".



To find the chances of an assualt, simply divide the number of recorded assualts (17) by the total number violent crimes (50) like so:


{{{P(robbery)=n(assualts)/n(violent_crimes)=17/50=0.34}}}


So there is a 34% chance that the next violent crime will be an assualt.


Note: simply multiply the decimal value 0.34 by 100 to get 34%



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# 3



In this problem, the population of Pennsylvannia is extra information. In other words, you don't use this info to find the answer (and it's there to confuse you).



Number of people who do NOT live in California = Number people in US - Number of people who live in California


Number of people who do NOT live in California = 293,655,404 - 35,893,799 = 257,761,605

So

Number of people who do NOT live in California = 257,761,605


"What is the probability that a randomly selected US resident did not live in California? "


So the probability of selecting a US resident that does NOT live in California is...


P(NOT California) = # who do NOT live in California/# total

P(NOT California) = 257,761,605/293,655,404

P(NOT California) = 0.877769


So the percentage is 87.7769% which rounds to 87.777% (to the nearest thousandth of a percent). 



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# 4


When they report the odds of 9:4, this means that the chances of losing is {{{9/13}}} and the chances of winning are {{{4/13}}} (note: {{{9+4=13}}} which is the total). These odds form the ratio



*[Tex \LARGE \frac{\text{Odds Against}}{\text{Odds For}}=\frac{\text{Winnings}}{\text{Amount Customer Paid}}]



and if we plug the given values in, we get the ratio: {{{9/4=x/8}}}



{{{9/4=x/8}}} Start with the given ratio



{{{(9/4)*8=x}}} Multiply both sides by 8



{{{72/4=x}}} Multiply



{{{18=x}}} Simplify



So our answer is {{{x=18}}} which means that Sally wins $18 (which makes the answer choice C)


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# 5


First, list out the probabilities for each possible event:


P(grand prize) = 1/1000

P(consolation prize) = 2/1000 = 1/500

P(losing) = 997/1000



Note: since there are 3 prize positions, there are 1000-3=997 losing positions.



Now simply multiply the probabilities with their corresponding winning value and then add up those products. In other words.... 


Expected Value = P(Grand Prize)*Value of Grand Prize + P(Consolation Prize)*Value of Consolation Prize + P(losing)*Value of losing 



Now plug in the given values and evaluate:



Expected Value = (1/1000)*(100) + (1/500)*(75) + (997/1000)*(0)


Expected Value = 1/10 + 3/20 + 0 


Expected Value = 5/20 


Expected Value = 1/4 


Expected Value = 0.25



So he expects to make $0.25. In other words, on an average try, he'll make 25 cents.


However, he spent $3 to play the game so this means that we need to subtract the cost to play from the expected winnings to get: 


Expected Average Winnings - Cost to Play = $0.25 - $3 = -$2.75


So he expects to lose $2.75 overall which means that the answer is A).


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# 6


To find the expected value, simply perform a weighted average:



Weighted Average:


{{{(Ss+Cc+Dd)/(total)}}} 


where 


S = Number of Suspense Movies
s = Number of Suspense Movie Viewers
C = Number of Comedy Movies
c = Number of Comedy Movie Viewers
D = Number of Drama Movies
d = Number of Drama Movie Viewers


So plug in the given values and evaluate:


{{{(15*3500 + 40*4500 + 45*6000)/100}}}



{{{(52500 + 180000 + 270000)/100}}}



{{{502500/100}}}



{{{5025}}}



So the expected number of viewers is 5,025 which means that the answer is D)


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# 7



Take note that the even and odd digits are

Even: 0, 2, 4, 6, 8
Odd: 1, 3, 5, 7, 9


So there are 5 even and 5 odd numbers.


Also, the number of values from 0 to 9 is 10.


So to find the number of combinations, simply multiply the number of choices for each digit. So multiply the number of choices for the first digit (5) by the number of choices for the second digit (10) by the number of choices for the third digit (5):


5*10*5=50*5=250


So 250 different numbers can be formed based on the given criteria.


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Side Note (ignore if you understand method given above): if you aren't sure about the multiplication, try using smaller numbers. So let's say we can only use the numbers 1, 2, and 4. How many 2 digit numbers can we form (with repeated values)?


Well, we can list them out: 

11, 12, 14,
21, 22, 24,
41, 42, 44


From this list, we can see that there are 9 possible numbers. Take note that we have 3 choices for each digit. So multiplying 3*3 gets us 9 which is the original answer. Also, you can draw a tree to visually reinforce the idea.


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# 8


Whenever we want to find the probability of event A OR event B happening, we simply add the two probabilities like so:


*[Tex \LARGE P(\text{A OR B})=P(\text{A})+P(\text{B})]



*[Tex \LARGE P(\text{Caramel OR Nuts})=P(\text{Caramel})+P(\text{Nuts})] ...  Start with the given probability



*[Tex \LARGE P(\text{Caramel OR Nuts})=\frac{\text{Number of Caramel Candies}}{\text{Total Number of Candies}}+\frac{\text{Number of Nut Candies}}{\text{Total Number of Candies}}] ... Expand



*[Tex \LARGE P(\text{Caramel OR Nuts})=\frac{7}{20}+\frac{9}{20}] ... Plug in the given values.



*[Tex \LARGE P(\text{Caramel OR Nuts})=\frac{7+9}{20}] ... Add the fractions.



*[Tex \LARGE P(\text{Caramel OR Nuts})=\frac{16}{20}] ... Add



*[Tex \LARGE P(\text{Caramel OR Nuts})=\frac{4}{5}] ... Reduce



So the chances are {{{4/5}}} which is 0.8 in decimal form which gives an 80% chance.



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# 9


a)


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/work/w5a39a-2.png">


Using the tree, we get the sample space:


{b,b,b,b}
{b,b,b,g}
{b,b,g,b}
{b,b,g,g}
{b,g,b,b}
{b,g,b,g}
{b,g,g,b}
{b,g,g,g}


{g,b,b,b}
{g,b,b,g}
{g,b,g,b}
{g,b,g,g}
{g,g,b,b}
{g,g,b,g}
{g,g,g,b}
{g,g,g,g}


So for instance {b,g,g,b} means that the couple had a boy, girl, girl, and then a boy (in that order).


Note: recall, the sample space is the set of ALL possible outcomes.

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b)

Since we want to know the chances of the couple having "at least three boys", this means they want to know the chances of having 3 boys OR 4 boys (since <i>at least</i> means that figure or more).



Looking back at the list of all possible outcomes (ie the sample space) from part a), we see that we have the combinations for 3 boys:

{b,b,b,g}, {b,b,g,b}, {b,g,b,b}, and {g,b,b,b}


So there are 4 cases where the couple would have 3 boys. 



So *[Tex \LARGE P(\text{3 boys ONLY})=\frac{\text{Number of Cases with 3 boys only}}{\text{Total number in sample space}}=\frac{4}{16}=\frac{1}{4}] 


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Also, since there is only ONE way to have 4 boys (of a total of 4 children), this means that
 
 
 
*[Tex \LARGE P(\text{\text{4 boys ONLY}})=\frac{\text{Number of Cases with 4 boys only}}{\text{Total number in sample space}}=\frac{1}{16}]



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Now simply add the two probabilities to find the chances of either one occurring:


*[Tex \LARGE P(\text{AT LEAST 3 boys})=P(\text{3 boys OR 4 boys})]



*[Tex \LARGE P(\text{AT LEAST 3 boys})=P(\text{3 boys ONLY})+P(\text{4 boys ONLY})]



*[Tex \LARGE P(\text{AT LEAST 3 boys})=\frac{\text{Number of Cases with 3 boys only}}{\text{Total number in sample space}}+\frac{\text{Number of Cases with 4 boys only}}{\text{Total number in sample space}}]



*[Tex \LARGE P(\text{AT LEAST 3 boys})=\frac{4}{16}+\frac{1}{16}]



*[Tex \LARGE P(\text{AT LEAST 3 boys})=\frac{4+1}{16}]



*[Tex \LARGE P(\text{AT LEAST 3 boys})=\frac{5}{16}]



So the probability of the couple having AT LEAST 3 boys is {{{5/16}}} which is 0.3125 in decimal form which gives a 31.25% chance


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# 10


a)


Since there are 12 face cards (there are 4 groups of jack, queen, and king cards) and there are 52 total cards, this means that there are {{{52-12=40}}} cards that are NOT face cards.


Since we know that the given card is NOT a face card, this means we can focus solely on the 40 non face cards. So the number of the sample space is 40. 


Also, since there are 4 cards labeled "7", this means that the number of elements in the event space is 4. 


So


*[Tex \LARGE P(\text{drawing a 7 \| NOT a face card})=\frac{\text{number in event space}}{\text{number in sample space}}=\frac{\text{number of 7 cards}}{\text{number of non face cards}}=\frac{4}{40}=\frac{1}{10}]



So the chances of drawing a 7 given that the card is NOT a face card is {{{1/10}}} which is 0.1 or 10%



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b)


Since spade cards are black and the given card is red, this means that the probability of choosing a spade given the card is red is 0%. In other words, it is IMPOSSIBLE to choose a spade card given the card is red.



More formally, we can say that



Number of spades that are red = 0
Number of Red Cards = 26



P(spade | red) = 0/26 = 0


So either way, we get the same answer of 0