Question

Ch:-Probability


In a single throw of two coins, find the probability of getting each of the following.
(1) Tails on both (i) Tails at least on one (iii) Tails on one (iv) No tails


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First Correct Answer will be Marked As Brainlist.....​

Answer 1

Step-by-step explanation:

I) p(tails on both)=favourable outcomes /total outcome

=1/4

ii) p(tails at least on one) =favouarble outcomes/total outcomes

=2/4

iii) p(no tails )=write formula

=1/4

Answer 2

$\underline{ \underline{ \bf\red{information}}}$
Total possible outcomes in a single throw of two coins = 4

The outcomes are :—
☺{H , H }
☺{H , T }
☺{ T , H }
☺{ T , T }

$\blue{ \bf option \: \red{1}}$

Tails on both, i.e.: { T, T }

No. of desired outcome = 1

Thus, P( tails on both )
$= \sf \dfrac{no. of \: favourable \: outcome}{total \: no. \: of \: outcome} = \green{\dfrac{1}{4} }$

$\bf \blue{option} \: \red{2}$

Tails at least on one, i.e.: { H, T } and { T, H } and { T, T }

No. of desired outcomes = 3

Thus, P ( tails at least on one )
$= \sf \dfrac{no. of \: favourable \: outcome}{total \: no. \: of \: outcome} = \green{\dfrac{3}{4} }$

$\bf \blue{option} \: \red{3}$

Tails on only one, i.e.: { H, T } and { T, H }

No. of desired outcomes = 2

Thus, P ( tails on only one )
$= \sf \dfrac{no. of \: favourable \: outcome}{total \: no. \: of \: outcome} = \dfrac{ \cancel{2}}{ \cancel{4}} = \green{\dfrac{1}{2} }$

$\bf \blue{option} \: \red{4}$

No tails, i.e.: { H, H }

No. of desired outcome = 1

Thus, P ( no tails )
$= \sf \dfrac{no. of \: favourable \: outcome}{total \: no. \: of \: outcome} = \green{\dfrac{1}{4} }$

$\mid \underline{\underline{\LARGE\bf\green{Brainliest \: Answer}}}\mid$

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