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Veronika [31]
3 years ago
7

David has a bag that holds 18 pens. 5 of the pens are ball point, 4 are gel pens and 9 are felt tip marker pens. He reaches into

the bag to pull out one pen. If every pen has an equal chance of being pulled out of davids bag, what is the probability that he will pull out a gel pen
Mathematics
2 answers:
sveticcg [70]3 years ago
8 0
If you take away 18-5= 13 then 13-9= 4 so thats the probability that it is 4

Flauer [41]3 years ago
8 0
The probability is 4
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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

• 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
2 years ago
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