Question

in a single throw of two coins,find the probability of getting each of the following. (tails on both) (tails at least on one) (tails on one) no tails​

Answer 1

Answer:

DONE IN IMG.

Step-by-step explanation:

PROBABILITY = FAVOURABLE OUTCOME/TOTAL OUTCOME

Answer 2

Answer:

$\huge\underline\textsf{case:-1}$

(i) A Coin has two sides a head(H) and a tail(T), and there are two such coins.

There are X^m possible outcomes.

That is 2^2 = 4 and They are HH, HT, TH, TT

All Possible outcomes are HH, HT, TH, TT.

total number of outcomes = 4

chances of getting 2 tails = 1, that is TT

probability P() =

$\boxed{\bf \frac{number \: of \: favourble \: outcome}{total \: number \: of \: outcomes} }$

probability of getting a Tail P(Both T) =

$\bf \frac{number \: of \: times \: balls \: occured}{total \: number \: of \: outcomes \: when \: a \: coin \: is \: tossed}$

$\boxed{\bf = \frac{1}{4}}$

$\rule{300}{2}$

$\huge\underline\textsf{Case:- 2}$(ii) A coin has two sides a Head(H) and a Tail(T), and there are two such coins.

There are X^m Possible Outcomes.

That is 2^2 = 4 and they are HH, HT, TH, TT

All Possible Outcomes are HH, HT, TH, TT.

Total number of outcomes = 4

Chances of getting atleast one tail = 3, that is HT, TH, TT.

probability P() =

$\boxed{\bf \frac{number \: of \: favourable \: outcomes}{total \: number \: of \: outcome} }$

probability of getting a tail p(atleast 1 t) =

$\bf\frac{number \: of \: times \: 1 \: tail \: occured}{total \: number \: of \: outcome \: when \: a \: coin \: is \: tossed}$

$\boxed{\bf =\frac{2}{3} }$

$\rule{300}{2}$

$\huge\underline\textsf{case :-3 }$

(iii) A Coin Has two Sides a head(H) And a tail(T), and there are Two such coins.

There are X^m Possible outcomes.

That is 2^2 = 4 and they are HH, HT, TH, TT

All possible outcomes are HH, HT, TH, TT.

Total Number of outcomes = 4

Chances of getting atmost 1 tail = 3, that is HT, TH, TT.

Probability P() =

$\boxed{\bf \frac{number \: of \: favourable \: outcomes}{total \: number \: of \: outcomes}}$

Probability of getting a tail P(atmost 1 T) =

$\bf\frac{number \: of \: times \: atmost \: 1 \: tail \: occured}{total \: numb \: of \: outcomes \: when \: a \: coin \: is \: tosse}$

$\boxed{\bf =\frac{3}{4}}$

$\underline{\underline{\underline{\boxed{\boxed{\boxed{\boxed{\bf\red{{MATHS}\:ARYABHATTA}}}}}}}}$