Three unbiased coins are tossed simultaneously. Find the probability of getting exactly ^@2^@ heads.


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Answer:

^@ \dfrac { 3 } { 8 } ^@

Step by Step Explanation:
  1. When ^@3 ^@ coins are tossed simultaneously, all the possible outcomes are ^@ HHH, HHT, HTH, THH, HTT, THT, TTH,^@ and ^@TTT. ^@

    So, total number of possible outcomes = ^@ 8 ^@.
  2. Let ^@ E ^@ be the event of getting exactly ^@ 2 ^@ heads.

    The favourable outcomes are ^@ HHT, HTH, ^@ and ^@ THH. ^@

    Number of favouable outcomes = ^@ 3 ^@ @^ \begin{aligned} \therefore \space P(\text{ getting exactly 2 heads }) = P(E) = \dfrac { \text { Number of favourable outcomes } } { \text { Total number of outcomes } } = \dfrac { 3 } { 8 }. \end{aligned} @^ Thus, the probability of getting exactly ^@ 2 ^@ heads = ^@ \dfrac { 3 } { 8 } ^@

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