Question

Three coins are tossed once. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) at most two tails.

Answer 1

It is given that three coins are tossed once, then, the sample space will be,

S={ HHH,HHT,HTH,THH,HTT,THT,TTH,TTT }

So, n( S )=8

(i)

Consider that B is the event of occurrence of 3 heads, then, B={ HHH } and accordingly, n( B )=1.

The probability will be,

P( B )= n( B ) n( S ) = 1 8

(ii)

Consider that C is the event of occurrence of 2 heads, then, C={ HHT, HTH, THH } and accordingly, n( C )=3.

The probability will be,

P( C )= n( C ) n( S ) = 3 8

(iii)

Consider that D is the event of occurrence of atleast 2 heads, then, D={ HHH, HHT, HTH, THH } and accordingly, n( D )=4.

The probability will be,

P( D )= n( D ) n( S ) = 4 8 = 1 2

(iv)

Consider that E is the event of occurrence of at most 2 heads, then, E ={ HHT, HTH, THH, HTT, THT, TTH, TTT } and accordingly, n( E )=7.

The probability will be,

P( E )= n( E ) n( S ) = 7 8

(v)

Consider that F is the event of occurrence of no head, then, F={ TTT } and accordingly, n( F )=1.

The probability will be,

P( F )= n( F ) n( S ) = 1 8

(vi)

Consider that G is the event of occurrence of 3 tails, then, G={ TTT } and accordingly, n( G )=1.

The probability will be,

P( G )= n( G ) n( S ) = 1 8

(vii)

Consider that H is the event of occurrence of exactly 2 tails, then, H={ HTT, THT, TTH } and accordingly, n( H )=3.

The probability will be,

P( H )= n( H ) n( S ) = 3 8

(viii)

Consider that I is the event of occurrence of no tail, then, I={ HHH } and accordingly, n( I )=1.

The probability will be,

P( I )= n( I ) n( S ) = 1 8

(ix)

Consider that J is the event of occurrence of at most 2 tails, then, J={ HHH, HHT, HTH, THH, HTT, THT, TTH } and accordingly, n( J )=7.

The probability will be,

P( J )= n( J ) n( S ) = 7 8