Question

If a coin is tossed three times, find probability of getting
1) Exactly 3 tails
2) Atmost 2 heads
3) Atleast 2 tails
4) Exactly 2 heads
5) Exactly 3 heads​

Answer 1

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

‣ 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

1) Exactly 3 tails

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

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

2) At most 2 heads

[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

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

3) At least 2 tails

[It means there can be two or more tails]

Here

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

• Total outcomes = 8

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

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

4) Exactly 2 heads

Here

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

• Total outcomes = 8

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

5) Exactly 3 heads

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

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

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