# A die is rolled once find the probability of getting (a) a prime number (b) number lying between 2and 6 (c) an odd number(d) a perfect square

2
by hrithiksagar

2016-02-01T21:09:55+05:30
Total no.of outcomes = 6
(a) prime numbers = 2,3,5
Probability of getting a prime number = 3/6 = 1/2
(b)numbers lying between 2 and 6 are 3,4,5
Probability of getting a number which is lying in between  2 and 6 is = 3/6 = 1/2
(c)odd numbers = 1,3,5
Probability of getting an odd number = 3/6 = 1/2
(d) perfect square numbers = 4
Probability of getting a perfect square number = 1/6

plzz mark it as the best if u like it
hye i have a doubt 1 is also a perfect square right then the last probability must be 2/6 =1/3
ok thankyou
you are welcome
2016-02-01T21:12:57+05:30
Total no.of outcomes = n(S) = 6
(i) Let E be the event of getting a prime number
n(E) = 3    [ 2,3,5 ]
p(E) = n(E)/n(S)
= 3/6
= 1/2
(ii) Let E be the event of getting a no lying between 2 and 6
n(E) = 3    [3,4,5]
p(E) = 3/6
= 1/2
(iii) Let E be the event of getting an odd no
n(E) = 3    [ 1,3,5 ]
p(E) =3/6
= 1/2
(iv) Let E be the event of getting a perfect square no
n(E) = 2   [ 1,4]
p(E) = 2/6
= 1/3