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MA_775_DIABLO [31]
3 years ago
10

Tony allows his customers to create their own pizza. The customers choose exactly one type of crust, sauce, cheese, and meat for

their pizza. The following table shows the options available to customers. How many different pizza combinations are there?​

Mathematics
2 answers:
Natalka [10]3 years ago
4 0

In cases like these, you simply have to multiply the choices you have for each step.

In fact, you can choose between 4 different crusts. For each of these crusts you can choose 4 different sauces. This means that there are 4\cdot 4 = 16 choices at this point. In fact, if we call the crust options 1,2,3,4 and the sauce options A,B,C,D, the 16 possible combinations are

1A,\ 1B,\ 1C,\ 1D,\ 2A,\ 2B,\ 2C,\ 2D,\ 3A,\ 3B,\ 3C,\ 3D,\ 4A,\ 4B,\ 4C,\ 4D

From here, you keep going multiplying the number of options for each step, for a total of

4\cdot 4 \cdot 3 \cdot 4 = 192

Goryan [66]3 years ago
3 0

Answer:

Straight up the answer is 192

Step-by-step explanation:

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If a coin is tossed three times, find probability of getting
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{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

‣ A coin is tossed three times.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

‣ The probability of getting,

1) Exactly 3 tails

2) At most 2 heads

3) At least 2 tails

4) Exactly 2 heads

5) Exactly 3 heads

{\large{\textsf{\textbf{\underline{\underline{Using \: Formula :}}}}}}

\star \: \tt  P(E)= {\underline{\boxed{\sf{\red{  \dfrac{ Favourable \:  outcomes }{Total \:  outcomes}  }}}}}

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

★ When three coins are tossed,

then the sample space = {HHH, HHT, THH, TTH, HTH, HTT, THT, TTT}

[here H denotes head and T denotes tail]

⇒Total number of outcomes \tt [ \: n(s) \: ] = 8

<u>1) Exactly 3 tails </u>

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

\therefore  \sf Probability_{(exactly  \: 3 \:  tails)}  =  \red{ \dfrac{1}{8}}

<u>2) At most 2 heads</u>

[It means there can be two or one or no heads]

Here

• Favourable outcomes = {HHT, THH, HTH, TTH, HTT, THT, TTT} = 7

• Total outcomes = 8

\therefore  \sf Probability_{(at \: most  \: 2 \:  heads)}  =  \green{ \dfrac{7}{8}}

<u>3) At least 2 tails </u>

[It means there can be two or more tails]

Here

• Favourable outcomes = {TTH, TTT, HTT, THT} = 4

• Total outcomes = 8

\longrightarrow   \sf Probability_{(at \: least \: 2 \:  tails)}  =  \dfrac{4}{8}

\therefore  \sf Probability_{(at \: least \: 2 \:  tails)}  =   \orange{\dfrac{1}{2}}

<u>4) Exactly 2 heads </u>

Here

• Favourable outcomes = {HTH, THH, HHT } = 3

• Total outcomes = 8

\therefore  \sf Probability_{(exactly \: 2 \:  heads)}  =  \pink{ \dfrac{3}{8}}

<u>5) Exactly 3 heads</u>

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

\therefore  \sf Probability_{(exactly \: 3 \:  heads)}  =  \purple{ \dfrac{1}{8}}

\rule{280pt}{2pt}

8 0
1 year ago
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