Question

Two dice are thrown simultaneously find the probability of getting (i) the sum as a prime number (ii) a doublet of even number

Answer 1

Heya Frnd ........☺

Total no. of outcomes = Square of tot. outcomes of 1 die

= 6² = 36

NOW,

★ For the sum to be prime no. , the last prime could be 31.

NOW,

Primes till 31

= 2,3,5,7,11,13,17,19,23,29,31.

NOW,

$P(E) \: = \: \frac{no. \: of \: favourable \: outcomes}{tot. \: no. \: of \: outcomes}$

THEREFORE,

11 / 36 ...........★ANS 1★

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★ DOUBLETS OF EVEN NO. = (2,2),(4,4),(6,6).

SO,

ANSWER = 3/36

= 1/12 ..............★ ANS 2 ★

________☺☺☺_________

HOPE IT WILL HELP U ...........✔

Answer 2

hey!
here's your answer:

n(s)=6×6
n(s)=36

let A be the event that the sum of the uppermost faces is a prime number.

A={(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4)
,(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)}
n(A)=14.

P(A)=n(A)/n(S)
P(A)=14/36
P(A)=7/18

Let B be the event that there is doublet of even number.

B={(2,2),(4,4),(6,6)}
n(B)=3

P(B)=n(B)/n(S)
P(B)=3/36
P(B)=1/12

the probability that the sum is a prime number is 7/18

and the probability of getting a doublet of even number is 1/12

hope this helps!!!