A box contains 4 red chips and 2 blue chips. If two chips are selected

Question

A box contains 4 red chips and 2 blue chips. If two chips are selected at random without replacement, what is the probability that the chips are different colors?

A. 1/2
B. 8/15
C. 7/12
D. 2/3
E. 7/10

............................................................
My solution:
Got the correct answer by the following method.
(4/6)(2/5)+(2/6)(4/5)= 8/15>>OA

My querie:
Why are we not getting the answer when we do 1-same color??
1-Both red + 1-both blue
(4/6)(3/6)+(2/6)(1/6)

feruz77 · Accepted Answer

Total selection ways: 6C2=6!/2!4!=15
Selecting one blue chip out of two: 2C1=2!/1!1!=2
Selecting one red chip out of four: 4C1=4!/1!3!=4
Thus, (2C1*4C1)/6C2=(2*4)/15=8/15

Bunuel · Answer

First solution is right.

As for the second one: you've made a little mistake you've avoided when solving for the different colors.

We surely can get the answer for same color by exactly the same method:

P(both same color)=4/6*3/ 5 +2/6*1/ 5

You just forgot that when taking the first chip there are 5 left, so the chances of getting is out of 5, not out of 6.

Hope it's clear.

tejal777 · Answer

thank you..I'll go jump off the balcony now..!!grumble grumble..Silly silly!!

mvictor · Answer

can someone explain if the method used by me is correct?

we have 6 chips, and out of which we have to select 2, hence 6C2=15
we have 4 red chips - 4C1=4
2 blue chips - 2C1=2
then we multiply the combinations 4x2 and get 8
putting all together, we have the probability of selecting one red and one blue - 8/15

mvictor · Answer

Bunuel, can u pls specify if my method is good to use for solving this problem

Bunuel · Answer

Yes, that's correct.

Gaurav9588 · Answer

(Just for my practice) I am considering this problem WITH replacement : * One (conventional) way to solve is ( x + x ) = ; * BUT, I am struggling hard to solve the same using Combinations : My approach demands a multiplication by 2

. Anyone to help? TIA !

adkikani · Answer

niks18   pushpitkc   chetan2u   pikolo2510

What is incorrect in below approach?

Probability = (Favorable outcomes) / (total no of outcomes)
 
P (red) = 1/4
P (blue) = 1/2
Combined probability = P(r)* P(b) = 1/8

chetan2u · Answer

P(r) is 4/6... Since 4 reds are there so favourable outcomes and 6 total outcomes..
Since it is without replacement, only 5 are left, so total outcomes =7 and blue are 2 so P(b)=2/5

If blue first then 2/6 and red 4/5

ylee6262 · Answer

Question...

Why is (4/6)(2/5) + (2/6)(4/5) correct answer?

The way I am interpreting this question is that whatever you draw on the first attempt does not matter.

If you get RED chip on the first draw, you just need to draw blue chip on the second draw, which is probability of 2/5.
If you get BLUE chip on the 1st draw, you just need to draw red chip on the 2nd draw, which is probability of 4/5.

Because what you draw on your 1st pick doesn't matter, wouldn't we just add (2/5) + (4/5)?

Please help!!

philipssonicare · Answer

ylee6262 
I find the direct approach confusing. Do it the other way and work out how to have all the same colour

Chance all red 4/6·3/5=2/5=6/15
Chance all blue 2/6·1/5=1/15

15/15-6/15-1/15=8/15

BrentGMATPrepNow · Answer

P(different colors) = P(1st chip is red   and   2nd chip is blue    OR    1st chip is blue   and   2nd chip is red)
= P(1st chip is red   and   2nd chip is blue)    +    P(1st chip is blue   and   2nd chip is red)
= P(1st chip is red)   x   P(2nd chip is blue)    +    P(1st chip is blue)   x   P(2nd chip is red)
= 4/6   x   2/5    +    2/6   x   4/5
= 8/30    +    8/30
= 16/30
= 8/15

Answer: B

Cheers,
Brent

ScottTargetTestPrep · Answer

To satisfy the requirement, we must obtain either R-B or B-R. Thus, we need to determine:

P(red) x P(blue) + P(blue) x R(red)

4/6 x 2/5 + 2/6 x 4/5 = 2/3 x 2/5 + 1/3 x 4/5 = 4/15 + 4/15 = 8/15.

Answer: B

mbaprep2020 · Answer

P(red) x P(blue) + P(blue) x R(red)

4/6 x 2/5 + 2/6 x 4/5 = 2/3 x 2/5 + 1/3 x 4/5 = 4/15 + 4/15 = 8/15.

HWPO · Answer

Why do we care about the OR option here? Why isn't it just one case of 4/6 * 2/5 ? Even if you get a blue chip first, it'll still be the same probability 2/6 * 4/5 I just don't understand why we need to account for the OR. The situation remains the same whether you get a blue or red chip first

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