Question

Please anyone answer it correctly i will follow you and mark as brainliest​

Answer 1

Step-by-step explanation:

3 coins tossed

therefore the possibilities of are

s={HHH,HTH,TTH,THH,HHT,HTT,TTH,THT}

n(s)=8

(1)probability of 3 tails

A={TTT}

n(A)=1

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

=1/8

(2)probability of 2 tails

B={TTH,HTT,THT}

n(B)=3

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

=3/8

(3)probability of getting no tails

C={HHH}

n(C)=1

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

= 1/8

(4) probability of getting 2heads nd 1tail

D={HTH,THH,HHT}

n(D)= 3

P(D)=n(D)/n(S)

=3/8

(5)probability of getting atleast one head

E={HTT,THT,TTH,HTH,THH,HHT,HHH}

n(E)=7

P(E)=n(E)/n(S)

=7/8

I hope this helps you!!!!!!

Answer 2

Answer:

v) at least one head

Step-by-step explanation:

As the probability of getting head =1/2

and the probability of getting tails =1/2