Problem Solving

1st draw 3 marbles can be selcted out of 9 in 9C3 = 84 ways

now here only prime nos are 2,3,5,7 rest of the 5 nos are not prime

so amongst these 84 combinations, the combinations which have exactly 1 prime are 4*5C2 = 40

so probabilty is 40/84 = 10/21

now probabilty that prme is selected while placing a ball in the cup is 1/3

so combined probabilty is 10/21 * 1/3 = 10/63

for next draw we can select 3 out of 6 in 6c3 = 20 ways

& we need 2 primes this time

so 2 primes out of 3 can be selected in 3 ways

& the other nos should be non prime i.e it can be selected out of the left 3 non prime nos in 3 ways

so together it is 3*3 =9 ways

so probabailty is 9/20

now probability that 2 primes are selected to place in the cup is 1/3

so combined P = 9/20 *1/3 = 3/20

last draw 3 balls out of 3 P=1

now probability that 1 prime is selected to put in the cup is 1/3

so combined Probability is 1*1/3 =1/3

so the reqd Probabilty is

10/63 * 3/20 * 1/3 = 1/126

Ans should be A