Question

varun throws a die find the probability of getting
(i) a number <4
(ii) a number <3
(iii) number 5​

Answer 1

Step-by-step explanation:

total probabilities= 6

probability $=\frac{probabilities that satisfy the condition}{total probabilities}$

i) numbers less than 4 in a dice (1,2,3)

probabilities that satisfy the condition= 3

$\frac{3}{6}=\frac{1}{2}$

1/2

ii)  numbers less than 3 in a dice (1,2)

probabilities that satisfy the condition= 2

2/6 = 1/3

iii)   numbers equal to 5 in a dice = (5)

probabilities that satisfy the condition= 1

1/6

Answer 2

Step-by-step explanation:

s={ 1,2,3,4,5,6}

n(s) = 6

1]. A= { 5,6}

n(A)= 2

2] B={ 2,1}

n(B)= 2

3]C={5}

n(C)=1

P(A)=n(A)/n(S)= 2/6=1/3

P(B)= n(B)/n(S)=2/6=1/3

P(C)= n(C)/n(S)=1/6

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