Question

NO LINKS!! NOT A MULTIPLE CHOICE. PART 4 Please help​

Answer 1

Answers are in bold

4a) 13/49

4b) 32/245

4c) 22/49

4d) 37/49

4e) 213/245

5) 0 and 1

6) They must sum to 1

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Step-by-step explanation:

4a)

Add up all the numbers in the frequency column.

50+65+32+38+60 = 245

Then divide the frequency for ground beef (65) over the total (245)

65/245 = 13/49 is the probability of getting ground beef

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4b)

Divide the frequency for bacon (32) over the total calculated earlier.

The probability of getting bacon is 32/245

The fraction cannot be reduced further since 32 and 245 don't have any factors in common other than 1.

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4c)

50 people bought boneless chicken and 60 bought pork tenderloin.

This sums to 50+60 = 110 people out of the 245

110/245 = 22/49 is the probability of someone getting either boneless chicken or pork tenderloin.

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4d)

60 bought pork tenderloin out of the 245 total. That leaves 245-60 = 185 who did not buy pork tenderloin

185/245 = 37/49 is the probability of someone not getting pork tenderloin.

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4e)

32 people bought bacon out of the 245 people total. This means 245-32 = 213 people did not buy bacon.

The probability of not buying bacon is therefore 213/245. The fraction cannot be reduced further.

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5)

Probabilities must always be between 0 and 1.

0 represents "the event is impossible and will never happen"

1 represents "the event is guaranteed to happen"

In other words, 0 represents a 0% chance of it happening, while 1 represents a 100% chance of it happening.

If x is the probability of some event, then
`0 \le x \le 1`

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6)

The probabilities must sum to 1.

In other words, all the probabilities must add to 1.

Like mentioned earlier, the value 1 in terms of probability represents 100%

It's like adding fractions of a pizza and those fractions must add up to a whole pizza. We cannot add past 100% because this is the highest we can go.

Answer 2

`\begin{tabular} c \cline{1-2} \bf Meat Type & \bf Frequency\\\cline{1-2} Boneless chicken & 50 pounds\\\cline{1-2} Ground beef & 65 pounds\\\cline{1-2} Bacon & 32 pounds\\\cline{1-2} Turkey sausage & 38 pounds\\\cline{1-2} Pork tenderloin & 60 pounds\\\cline{1-2} \bf Total & \bf 245 pounds\\\cline{1-2}\end{tabular}`

`\sf Probability\:of\:an\:event\:occurring = (Number\:of\:ways\:it\:can\:occur)/(Total\:number\:of\:possible\:outcomes)`

Part (a)

`\textsf{P(Ground beef)}=\sf (65)/(245)=(13)/(49)`

Part (b)

`\textsf{P(Bacon)}=\sf (32)/(245)`

Part (c)

`\textsf{P(Pork tenderloin)}=\sf (60)/(245)`

`\textsf{P(Boneless chicken)}=\sf (50)/(245)`

`\implies \textsf{P(Pork tenderloin) or P(Boneless chicken)}=\sf (60)/(245)+(50)/(245)=(110)/(245)=(22)/(49)`

Part (d)

`\begin{aligned}\textsf{P(not Pork tenderloin)} & =1-\textsf{P(Pork tenderloin)}\\& = \sf 1-(60)/(245)\\& = \sf (37)/(49)\\\end{aligned}`

Part (e)

`\begin{aligned}\textsf{P(not Bacon)} & =1-\textsf{P(Bacon)}\\& = \sf 1-(32)/(245)\\& = \sf (213)/(245)\\\end{aligned}`