Question

i) If two coins are tossed, find the probability of (1) getting no tail (2) getting
no head (3) getting at least one tail. ans fast pls​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

We know,

$\rm\:Probability \: of \: an \: event =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes \: in \: sample \: space}$

Or

$\rm : \implies \: P(E) \: = \: \dfrac{n(E)}{n(S)}$

where,

• P(E) = probability of event
• n(E) = number of elements in favourable outcomes
• n(S) = number of elements in total outcomes.

Now,

When two coins are tossed, the possible outcomes are

• S = {(H,H), (H,T), (T,H), (T,T)}

This implies, number of elements in Samlpe space is

• n(S) = 4.

Answer :- 1.

Let E be the event getting no tail.

• So, E = {(H,H)}

This implies, number of elements in favourable outcomes

• n(E) = 1

So, required probability is

$\rm: \implies \: P(E) \: = \: \dfrac{1}{4}$

Answer :- 2.

Let E be the event getting no head.

• So, E = {(T,T)}

This implies, number of elements in favourable outcomes

• n(E) = 1

So, required probability is

$\rm : \implies \: P(E) \: = \: \dfrac{1}{4}$

Answer :- 3.

Let E be the event getting no tail.

• So, E = {(T,H), (H,T), (T,T)}

This implies, number of elements in favourable outcomes

• n(E) = 3

So, required probability is

$\rm : \implies \: P(E) \: = \: \dfrac{3}{4}$

Explore more :-

• The sample space of a random experiment is the collection of all possible outcomes.
• An event associated with a random experiment is a subset of the sample space.
• The probability of any outcome is a number between 0 and 1.
• The probability of sure event is 1.
• The probability of impossible event is 0.
• The probabilities of all the outcomes add up to 1.
• The probability of any event A is the sum of the probabilities of the outcomes in A.

Answer 2

Step-by-step explanation:

P = 1 /4

P=3 /4

PLEASE MARK AS BRAINLIEST