Question

Three unbiased coins are tossed together. find the probability of getting at least two heads is
1/2
1/8
3/8
1​

Answer 1

$\large\underline{\sf{Solution-}}$

The sample space when 3 coins are tossed is

$\rm :\longmapsto\:S = \{HHH, HHT,HTH,THH,HTT,THT,TTH,TTT\}$

$\bf\implies \:n(S) = 8$

Let assume that

E is the event of getting atleast 2 heads.

So, favourable outcomes are as follow :

$\rm :\longmapsto\:E = \{HHH, HHT,HTH,THH\}$

$\bf\implies \:n(E) = 4$

We know,

$\bf \:Probability\: of\: an \: event =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes \: in \: sample \: space}$

So,

$\rm :\longmapsto\:Probability\: of\: an \: event =\dfrac{n(E)}{n(S)}$

$\rm :\longmapsto\:Probability\: of\: an \: event =\dfrac{4}{8}$

$\rm :\longmapsto\:Probability\: of\: an \: event =\dfrac{1}{2}$

• So, option 1 is correct

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Explore more :-

The sample space of a random experiment is the collection of all possible outcomes.

The probability always lies between [0, 1].

The probability of sure event is 1.

The probability of impossible event is 0.