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7. A large cooler contains the following drinks: 5 lemonades, 9 Sprites, 7 Cokes, and 10 root beers. You randomly pick two cans, one at a time (without replacement). Compute the following probabilities.(a) What is the probability that you get two cans of Sprite? (b) What is the probability that you do not get two cans of Coke? (c) What is the probability that you get either two root beers or two lemonades? (d) What is the probability that you get one can of Coke and one can of Sprite? (e) What is the probability that you get two drinks of the same type?

7 A Large Cooler Contains The Following Drinks 5 Lemonades 9 Sprites 7 Cokes And 10 Root Beers You Randomly Pick Two Cans One At A Time Without Replacement Comp class=

Sagot :

A large cooler contains the following drinks,

5 Lemonades, let L reprersent Lemonade

9 Sprites, let S reprersent Sprite

7 Cokes, let C represent Coke and

10 Root beers, let R represent Root beer

Total drinks in the cooler is

[tex]=5+9+7+10=31[/tex]

Total outcome = 31 drinks

The formula of probability is

[tex]\text{Probability}=\frac{required\text{ outcome}}{total\text{ outcome}}[/tex]

a) The probability that you get two cans of Sprite is

[tex]\begin{gathered} Prob\text{ of picking the first Sprite without replacement is} \\ P(S_1)=\frac{9}{31} \\ Prob\text{ of picking the second Sprite is} \\ P(S_2)=\frac{8}{30} \\ \text{Probability of getting two cans of Sprite}=(PS_1S_2)=\frac{9}{31}\times\frac{8}{30}=\frac{12}{155} \\ (PS_1S_2)=\frac{12}{155} \end{gathered}[/tex]

Hence, the the probability that you get two cans of Sprite is 12/155

b)

The probability that you get two cans of Coke

[tex]\begin{gathered} Prob\text{ of picking the first Coke without replacement is} \\ P(C_1)=\frac{7}{31} \\ Prob\text{ of picking the second Coke is} \\ P(C_2)=\frac{6}{30} \\ \text{Probability of getting two cans of Coke is} \\ P(C_1C_2)=\frac{7}{31}\times\frac{6}{30}=\frac{7}{155} \\ P(C_1C_2)=\frac{7}{155} \end{gathered}[/tex]

The probability that you do not get two cans of Coke will be

[tex]\begin{gathered} \text{Prob that you do not get two cans of Coke is} \\ 1-P(C_1C_2)=1-\frac{7}{155}=\frac{155-7}{155}=\frac{148}{155} \\ \text{Prob that you do not get two cans of Coke }=\frac{148}{155} \end{gathered}[/tex]

Hence, the probability that you do not get two cans of Coke is 148/155

c)

The probability that you get two cans of Root beers is

[tex]\begin{gathered} Prob\text{ of picking the first Root b}eer\text{ without replacement is} \\ P(R_1)=\frac{10}{31} \\ Prob\text{ of picking the second Root b}eer\text{ is} \\ P(R_2)=\frac{9}{30} \\ \text{Probability of getting two cans of Root b}eer\text{ is} \\ P(R_1R_2)=\frac{10}{31}\times\frac{9}{30}=\frac{3}{31} \\ P(R_1R_2)=\frac{3}{31} \end{gathered}[/tex]

The probability that you get two cans Lemonades is

[tex]\begin{gathered} Prob\text{ of picking the first Lemonade without replacement is} \\ P(L_1)=\frac{5}{31} \\ Prob\text{ of picking the second Root b}eer\text{ is} \\ P(L_2)=\frac{4}{30} \\ \text{Probability of getting two cans of Lemonade is} \\ P(L_1L_2)=\frac{5}{31}\times\frac{4}{30}=\frac{2}{93} \end{gathered}[/tex]

The probability that you get either two root beers or two lemonades is

[tex]P(R_1R_2)+P(L_1L_2)=\frac{3}{31}+\frac{2}{93}=\frac{11}{93}[/tex]

Hence, the probability that you get either two root beers or two lemonades is 11/93

d)

[tex]\begin{gathered} Prob\text{ of picking the first Coke without replacement is} \\ P(C)=\frac{7}{31} \\ \text{Prob of picking a can of Sprite is} \\ P(S)=\frac{9}{30} \end{gathered}[/tex]

After getting both Sprite and Coke you will multiply the probabilities and then multiply them with 2 because you may choose Coke in first try and Sprite in second or the other way around

The probability that you get one can of Coke and one can of Sprite is

[tex]P(CandS)=2\times(\frac{7}{31}\times\frac{9}{30})=2(\frac{21}{310})=\frac{21}{155}[/tex]

Hence, the probability that you get one can of Coke and one can of Sprite is 21/155

e)

Prob of two of each of the cans of drinks (without replacement) are as follow

[tex]\begin{gathered} P(L_1L_2)=\frac{2}{93} \\ (PS_1S_2)=\frac{12}{155} \\ P(C_1C_2)=\frac{7}{155} \\ P(R_1R_2)=\frac{3}{31} \end{gathered}[/tex]

The probability that you get two drinks of the same type is

[tex]\begin{gathered} \text{Prob of two drinks of the same type is} \\ =P(L_1L_2)+(PS_1S_2)_{}+P(C_1C_2)+P(R_1R_2) \\ =\frac{2}{93}+\frac{12}{155}+\frac{7}{155}+\frac{3}{31}=\frac{112}{465} \\ \text{Prob of two drinks of the same type}=\frac{112}{465} \end{gathered}[/tex]

Hence, the probability that you get two drinks of the same type is 112/465

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