Hey Bunuel,
Could you please clarify the below doubt.
I know that we can calculate the probability by using Combination approach or by using Probability Approach.
The only doubt that I have is why do we need to multiply by 2 when I use the Probability approach.
i.e
\(P(RW)=
2*\frac{8}{8+y}*\frac{y}{8+y-1}=2*\frac{8}{8+y}*\frac{y}{7+y}\)
Is it because we are considering cases over here
i.e RW can occur in 2 ways - (R and then W) or (W and then R).
If yes, then when we calculate the probability(RW) by using combination approach.. we don't multiply the expression by 2. Why.
P(RW) = [(8C1)(yC1)]/[(8+y)C2]
Just to make it a bit clearer-
Question #1: A bag has 6 red marbles and 4 marbles. What are the chances of pulling out a red and blue marble.
Answer: (Chance of picking red * chance of picking blue) + Chance of picking blue*chance of picking red)
6/10*4/9 + 4/10*6/9 = 48/90
***Here, it appears that order matters, because you have to find the probability of getting red-blue, and add it to the probability of getting blue-red.
Question #2:
A bag has 4 red marbles, 3 yellow, and 2 green, what is the probability of getting 2 Red, 2 Green, and 1 yellow, if the marbles aren't replaced:
# of ways to get 2R, 2G, 1Y 2: 4C2*2C2*3C1 =18
Total # of ways to pick 5 balls: 9C5= 126
Answer: 18/126
Now here, it appears order DOES NOT matter, since we're using combinations (instead of permutations).
How come in Q#1, blue-red, red-blue are distinct, which means you have to add the probabilities of each together, but in Q#2, it uses combinations, which means order doesn't matter.
WHY? Completely lost
Please throw some light on this.
Thanks
-H
Bunuel wrote:
Orange08 wrote:
A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?
(1) y ≤ 8
(2) y ≥ 4
# of total marbles in the jar equals to \(8+y\), out of which \(R=8\) and \(W=y\)
\(P(RR)=\frac{8}{8+y}*\frac{8-1}{8+y-1}=\frac{8}{8+y}*\frac{7}{7+y}\);
\(P(RW)=
2*\frac{8}{8+y}*\frac{y}{8+y-1}=2*\frac{8}{8+y}*\frac{y}{7+y}\), multiplying by 2 as RW can occure in two ways RW or WR;
Question: is \(\frac{8}{8+y}*\frac{7}{7+y}>2\frac{8}{8+y}*\frac{y}{7+y}\)? --> is \(\frac{7}{2}>y\)? --> is \(y<3.5\)? eg 0, 1, 2, 3.
(1) y ≤ 8, not sufficient.
(2) y ≥ 4, sufficient, (\(y\) is not less than 3.5).
Answer: B.