Question

If three identical coins are tossed,
(a) What is the probability of getting all three heads?
(b) What is the probability of getting one head and two tails​

Answer 1

Answers

a). let E1 = ( HHH)

n(E1. =1

p(get) = p(E1) =n(E1) / ns = 1/8

b. let E2= one head

E2 =(HTT, THT, TTH)

n(E2) =3

p(get) = p(E2) = n(E2)/n(s) = 3/8

Answer 2

• The probability of getting all three heads = 1/8

• The probability of getting one head and two tails = 3/8

Given :

Three identical coins are tossed

To find :

• The probability of getting all three heads

• The probability of getting one head and two tails

Solution :

Step 1 of 3 :

Find total number of possible outcomes

Here it is given that three identical coins are tossed

So the outcomes are (H,H,H),(H,T,H),(T,T,T),(T,H,H),(T,T,H),(H,T,T),(T,H,T)

So total number of possible outcomes = 8

Step 2 of 3 :

Find total probability of getting all three heads

Let A be the event that of getting all three heads

So the event point for the event A is (H,H,H)

So total number of possible outcomes for the event A is 1

Hence the required probability of getting all three heads

= P(A)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: A }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{1}{8} }$

Step 3 of 3 :

Find the probability of getting one head and two tails

Let B be the event that of getting one head and two tails

So the event point for the event B is (T,T,H),(H,T,T),(T,H,T)

So total number of possible outcomes for the event B is 3

Hence the required probability of getting all three heads

= P(B)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: B }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{3}{8} }$

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