Question

hello brainly user !!

it's a question based on iit- jee 2014 - 1979 .



a pair of fair dice is rolled together till a sum of either 5 or 7 is obtained . then the probability that 5 comes before 7 is .....


fill in the blanks .

totally point !!! care don't fast answer first read question.!!!

Answer 1

⭐ hey mate ⭐





5 can be thrown in 4 ways and 7 can be thrown in 6 ways hence number of ways of throwing neither 5 nor 7 is


36 - ( 4 + 26) = 26

so,

probability of throwing a five in a single throw with a pair of dice is 4/36 = 1/ 9 and probability of throwing neither 5 nor 7 is 26/36 = 13/18.


hence required probability

$( \frac{1}{9} ) + ( \frac{13}{18} ) \: ( \frac{1}{9} ) + ({ \frac{13}{18} }^{2} )( \frac{1}{9} )$

1 / 9 / 1 - 13 / 18


= 2/5


be brainly

Answer 2

My approach :

Probability P(A) of getting 5 as sum on two dice is

P(A)=436=19

Let B the event that a sum of 7 occurs and C the event that neither a sum of 5 nor a sum of 7 occurs. We have :

P(B)=636=16

P(C)=2636=1318

Please suggest how to proceed further.