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3241004551 [841]
2 years ago
11

Walt has a box of 80 postage stamps. The box contains 16 stamps featuring the Bald Eagle and 40 stamps featuring the Stars and S

tripes. If Walt randomly chooses a stamp to paste on an envelope, what is the probability that the stamp features the Bald Eagle or the Stars and Stripes?
Mathematics
1 answer:
Sophie [7]2 years ago
6 0
7/10



16 + 40 is 56, so 56 out of 80 stamps in the box feature the Bald Eagle or the Stars and Stripes.

We can simplify 56/80 further though. Let’s divide the numerator and denominator each by 8 to get 7/10.

There’s your simplified answer.



Please consider marking this answer as Brainliest to help me advance.
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Step-by-step explanation:

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If a coin is tossed three times, find probability of getting
Assoli18 [71]

{\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

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

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

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Here

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

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

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[It means there can be two or more tails]

Here

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

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

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

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Here

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

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Here

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

\rule{280pt}{2pt}

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