Question

to coin are tossed simultaneously . find the probability of getting two head , at least one head and no head​

Answer 1

Answer:

S={HH,HT,TH,TT}.n(S)=4

let,

A be the event that getting two head.

A={HH}.n(A)=1

P(A)=n(A)/n(S)

=1/4

B be the event that getting at least one head.

B={HH,HT,TH}.n(B)=3

P(B)=n(B)/n(S)

=3/4

C be the event that getting no head.

C={TT}. n(C)=1

P(C)=n(C)/n(S)

=1/4.

I hope it's help you.

Answer 2

Answer:

Given :

Two unbiased coins are tossed simultaneously.

To find :

At least one head

No head

Solution :

When two unbiased coins are tossed, then its total outcomes :-

Total number of outcomes = HH, TT, HT, TH

As we know that

$\bullet\:{\boxed{\pmb{\frak{\purple{Probability =\dfrac{ Favourable\: outcomes}{Total\:number\:of\: outcomes}}}}}}$

As per given condition in the question

Probability of getting at least one head

★ Number of favourable outcomes = HH, HT, TH

★ Total number of outcomes = 4

Probability of getting at least one head = 3/4

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Probability of getting no heads

★ Number of favourable outcome = TT

★ Total number of outcomes = 4

Probability of getting no heads = 1/4

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