Question

What is the probaibility of getting product less than 18 when 2 dices are thrown simultaneously ?

Answer 1

Answer:

Step-by-step explanation:

S={36 no.}

n(S)= 36

Event

A ={1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4

5,1 5,2 5,3

6,1 6,2 6,3}

n(A)= 27

P(A)=n(A)

_ = 27 ÷36

n(S)

1:3

Answer 2

✝✝hey buddyy⬆⬆⬆⬆⬆⬆⬆

Given= product of the individual events must be less than 18..

hence,,,,,

sample space for the event is.......

S = (1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

from the sample space,,

set containing product less than 18,,, is

= {((1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6) (3,1)(3,2)(3,3)(3,4)(3,5)(4,1)(4,2)(4,3)(4,4)(5,1)(5,2)(5,3)(6,1)(6,2)}
so,,
N(s) = E= 26,,

since, n(s) = 36,,

therefore..

P(E) = $\frac{n(E)}{n(s)}$

P(E) = $\frac{26}{36}$

P(E) = $\frac{13}{18}$

$\huge{BE\:BRAINLY}$