For the experiment of tossing a coin three times, how do you find the probability of getting at least two heads?

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Answer 1 · 2017-08-22T08:17:25

$\frac{1}{2}$ $1/2$

Since we only have three throws it will be easier to list all possibilities, of 1st, 2nd and 3rd throws

$H =$ $H =$ head

$T =$ $T=$ Tail

$H H H$ $color(blue)(HHH)$

$H H T$ $color(blue)(HHT)$
$H T H$ $color(blue)(HTH)$
$T H H$ $color(blue)(THH)$

$T T H$ $T T H$
$T H T$ $THT$
$H T T$ $H T T$

$T T T$ $T T T$

Eight possibilities in all, and the requirement of at least two heads is shown in blue

$X = number of heads$ $X=" number of heads"$

$P (X \geq 2) = \frac{4}{8} = \frac{1}{2}$ $P(X >=2)=4/8=1/2$

Answer 2 · 2017-08-22T08:19:38

1/2

total possible case :- HHH, HHT, HTH, THH, TTT, TTH, THT, HTT

At least two heads :- 4

so the probability of getting at least two head is 4/8 = 1/2