rohansherry wrote:
Guys answer is A:
lets work out:
i cant explain the full thing but pls bear wid me...
prob = 1 - (prob selcting 1st no*prob not selcting 2nd no from the list) - ( prob selcting 2nd no*prob not selcting 1st no from the list) - (prob NOT selcting 1st no*prob NOT selcting 2nd no from the list)
NOTE: its 1 - ( p1 + p2 + p3 ) = 1- p1 -p2 -p3
now 15/45 = 1/3
prob = 1 - 1/3 * 14/44 - 1/3 * 14/44 - 30/45 *30/ 44
prob = 1/3
Pls if i am correct , dont hasitate in giving me my share lol........
if not then lets wait for the right solution.....cheers
Your approach is correct but calculation of probabilities is wrong. going your way, the result is still 7/66.
Define 4 possible outcomes of selecting 2 rocks.
P(S, NS) - probability to select 1st slate and 2nd nonslate
P(S,S) - 1st slate and 2nd slate
P(NS,S) - 1st non slate and 2nd slate
P(NS,NS) - 1st non slate and 2nd non slate
P(S,NS)+P(S,S)+P(NS,S)+P(NS,NS)=1
To find P(S,S), you have 2 approaches
1) Direct one as I showed before. \(\frac{1}{3}\times\frac{14}{44}\)
2) Not direct:
1-(P(S,NS)+P(NS,S)+P(NS,NS))=P(S,S)
P(S, NS) - \(\frac{1}{3}\times\frac{30}{44}\)
P(NS,S) - \(\frac{2}{3}\times\frac{15}{44}\)
P(NS,NS) -\(\frac{2}{3}\times\frac{29}{44}\)
p(S,S) = \(1-(\frac{1}{3}\times\frac{30}{44}+\frac{2}{3}\times\frac{15}{44}+\frac{2}{3}\times\frac{29}{44})=1-\frac{118}{132}=\frac{7}{66}\)
It is can be visualized by drawing a binary tree. It is always helpful not to get confused with conditional probs.
Sorry for edits...trying to figure out how to write formulas