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Sagot :
A fiar coin is tossed 3 times in succession.
The results for each experiment is displayed as follows;
[tex]\begin{gathered} \text{HHH} \\ \text{HHT} \\ \text{HTH} \\ \text{THH} \\ \text{HTT} \\ \text{THT} \\ \text{TTH} \\ \text{TTT} \end{gathered}[/tex]On each toss from the above results, the probability of getting a tail would include all results that has a tail come up. That would be;
[tex]P\lbrack\text{Event\rbrack}=\frac{\text{Number of required outcomes}}{Number\text{ of all possible outcomes}}[/tex][tex]P\lbrack\text{tail\rbrack}=\frac{4}{8}[/tex]Note that to get a tail on the "second toss" would mean to get a result with a tail as the second out of three. We have 7 outcomes with tails. However 4 of these has a tail as a second outcome, hence we have the required outcome as 4 out of a total of 8.
ANSWER:
Probability of getting a tail on the second toss is
[tex]P\mleft\lbrace\text{tail}\mright\rbrace=\frac{1}{2}[/tex]
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