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A fair coin is tossed 4 times. Provided that at least 2 heads occurred within the 4 tossings, what's the probability that at least one tail also occurred?

Ashutosh Pandey
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  • How many equally probably ways of getting at least two heads? How many of these do not have any tails? – Henry Aug 12 '20 at 15:27
  • This should be a routine application of definitions of conditional probability. Where could you have gotten stuck? – JMoravitz Aug 12 '20 at 15:38
  • If you really have no idea how to start, just list the $16$ possible outcomes and count. – lulu Aug 12 '20 at 15:44

1 Answers1

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$$ \begin{array}{l} \begin{array}{cccc} H & H & H & H \end{array} \\ \left.\begin{array}{cccc} H & H & H & t \\ H & H & t & H \\ H & t & H & H \\ t & H & H & H \end{array} \right\} \text{ 3 heads in 4 tosses} \\ \left.\begin{array}{cccc} H & H & t & t \\ H & t & H & t \\ t & H & H & t \\ H & t & t & H \\ t & H & t & H \\ t & t & H & H \end{array} \right\} \text{ 2 heads in 4 tosses} \end{array} $$ There are $6$ ways to get $2$ heads and $4$ ways to get $3$ heads, and just one way to get $4$ heads. Thus $11$ ways to get at least $2$ heads. In $10$ of those $11$ ways, there is at least $1$ tail.

Michael Hardy
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