Question

A coin and a die are tossed. State sample space and following events.
a) A: getting a head and an even number
b) B: getting a prime number
c) C: getting a tail and perfect square

Answer 1

s={T1,T2,,T3,T4,T5,T6,H1,H2,H3,H4,H5,H6}n(S)=12
A={H2,H4,H6}n(A)=3
B={T2,T3,T5,H2,H3,H5}n(B)=6
C={T1,T4}n(C)=2
hope it helps
plz mark brainliest

Answer 2

Answer:

$a)=\dfrac{1}{4}$

$b)=\dfrac{1}{2}$

$c)=\dfrac{1}{6}$

Step-by-step explanation:

In a coin there are a head and a tail. In a die there are 1 to 6 numbers.

When a coin and a die is tossed there are following sample space:

Total sample space={H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}

N(S)=12

Total cases=12

$Probability=\dfrac{ \textrm{Number of favorable cases}}{\textrm{Number of total cases}}$

We have to find probability of the following:

(a)A:Probability of getting a head and an even numbers:

Favorable cases(H2,H4,H6)=3

$Probability(A)=\dfrac{3}{12}=\dfrac{1}{4}$

(b)B:Probability of getting a prime number:

Favorable cases(H2,H3,H5,T2,T5,T6)=6

$Probability(B)=\dfrac{6}{12}=\dfrac{1}{2}$

(c)C:Probability of getting a tail and perfect square:

Favorable cases(T1,T4)=2

$Probability(C)=\dfrac{2}{12}=\dfrac{1}{6}$