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Three different coins are tossed together. Find the probability of getting:
$( i)$ exactly two heads
$( ii)$ at least two heads
$( iii)$ at least two tails.
Given: Three different coins are tossed together.
To do: To Find the probability of getting
$( i)$. exactly two heads
$( ii)$. at least two heads
$( iii)$. at least two tails.
Solution:
When three coins are tossed together, the possible outcomes are
$( HHH,\ HTH,\ HHT,\ THH,\ THT,\ TTH,\ HTT,\ TTT)$
$\therefore$ Total number of possible outcomes $=\ 8$
$( i)$. Favorable outcomes of exactly two heads are $( HTH,\ HHT,\ THH)$
$\therefore$ Total number of favourable outcomes $= 3$
$\therefore \ P( exactly\ two\ heads) =\frac{no.\ of\ favourble\ outcomes}{no.\ of\ total\ outcomes} =\frac{3}{8}$
$( ii)$. Favourable outcomes of at least two heads are $( HHH,\ HTH,\ HHT,\ THH)$
$\therefore$ Total number of favourable outcomes $=4$
$\therefore \ P( at\ least\ two\ heads) \ =\frac{no.\ of\ favourble\ outcomes}{no.\ of\ total\ outcomes}=\frac{4}{8}=\frac{1}{2}$
$( iii)$. Favourable outcomes of at least two tails are $( THT,\ TTH,\ HTT,\ TTT)$
$\therefore$ Total number of favourable outcomes$=4$
$\therefore \ P( at\ least\ two\ tails) =\frac{no.\ of\ favourble\ outcomes}{no.\ of\ total\ outcomes}=\frac{4}{8}=\frac{1}{2}$
Updated on: 10-Oct-2022
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