5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn WITHOUT replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?Given: 5B+3R+4P=12 marbles. Question: what is the probability of NOT having scenario BBPP.
Calculate the probability of an opposite event and subtract this value from 1.
Opposite event is that out of 4 marbles we will have 2 blue and 2 purple marbles (scenario BBPP):
\(P(opposite \ event)=\frac{4!}{2!2!}*\frac{5}{12}*\frac{4}{11}*\frac{4}{10}*\frac{3}{9}=\frac{4}{33}\), we are multiplying by \(\frac{4!}{2!2!}\) as scenario BBPP can occur in several ways: BBPP, PPBB, BPBP, ... So scenario BBPP can occur in \(\frac{4!}{2!2!}\) # of ways (# of permuations of 4 letters BBPP out of which 2 B's and 2 P's are identical);
Or with combinatorics \(P(opposte \ event)=\frac{C^2_5*C^2_4}{C^4_{12}}=\frac{4}{33}\).
So \(P=1-\frac{4}{33}=\frac{29}{33}\).
5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn WITH replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?Again Calculate the probability of an opposite event and subtract this value from 1.
Opposite event is that out of 4 marbles we will have 2 blue and 2 purple marbles:
\(P(opposite \ event)=\frac{4!}{2!2!}*\frac{5}{12}*\frac{5}{12}*\frac{4}{12}*\frac{4}{12}=\frac{25}{216}\), the same reason to multiply by \(\frac{4!}{2!2!}\);
So \(P=1-\frac{25}{216}=\frac{191}{216}\).
Similar problem:
probability-colored-balls-55253.htmlHope it helps.