Question

A pair of dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is...

Answer 1

@experimentX another question...

Answer 2

Shall I show what i did?

Answer 3

Okay

Answer 4

the probability that 5 comes before 7 is 4/10=2/5

Answer 5

Solution please..@Shriya

Answer 6

sample set= [(1,4), (2,3), (3,2), (4,1), (1,6) (2,5), (3,4), (4,3), (5,2), (6,1)]
so, n(S), i.e., the total number of elements in S=10
out of this, the first four terms show the sum=5.
therefore n(5)=4.
therefore, P(5)=n(5)/n(s)
therefore, P(5)=4/10=2/5

Answer 7

No..this can't be it..the sample space does not contain the right number..it should be 36 for two dice, whatever may be the condition

Answer 8

I got the same answer but in a different way...but I don't think your method is right

Answer 9

ok u calculated it considering the entire sample space. i considered just the points those are required. whatever be the criteria, the ratio remains the same. n hence our answers are mathing!

Answer 10

Yea..I got you now..that's right..Thanx for a new method !!

Answer 11

wc :)

Answer 12

There are 4 ways of getting a sum of 5 and 6 ways of getting a sum of 7. The sample space has 36 outcomes.
The probability of getting a sum of 5 or a sum of 7 is
\[P(5or7)=\frac{4}{36}+\frac{6}{36}=\frac{10}{36}\]
\[P(5first)=\frac{P(5)}{P(5or7)}=\frac{4}{36}\times \frac{36}{10}=0.4\]

Answer 13

Thank you @kropot72 for your answer..

Answer 14

You're welcome :)