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It's because 1 from each basket.

6 options from basket 1 and 8 options from basket 2
Total 6×8

It's 6C1 × 8C1
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WA@wfx1059

In Q145, why is total number of outcomes 6*8 =48 and not 14C2 since we will be selecting 2 fruits in total from the 14 fruits from the 2 baskets?
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This question is discussed in detail here: https://gmatclub.com/forum/carol-purcha ... 21728.html Hope it helps.
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WA@jit2975
Hey anyone who watched the recent Probability lecture from Aditya?
I have a doubt.

Would love to discuss.
Please ping
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I am solving this question, using the Permutation method taught by Aditya but my answer in 2 of them seems to be double of the actual answer. I am not sure what mistake I am making.

There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that:
a) all the balls chosen are different colours?
b) 2 of the balls are the same colour and other 2 are of different color each?
c) 2 of the balls are the same colour and other 2 are of different color each?
d) 3 of the balls are the same colour?

Assuming we have balls of colors A,B,C,D and E
a) P(ABCD)= (5/5)x(4/5)x(3/5)x(2/5) = 24/125 ✅
b) P(AABC) = (5/5)x(1/5)x(4/5)x(3/5)x4!/2! = 144/125 ❌. Answer is half of this i.e. 72/125
c) P(AABB) = (5/5)x(1/5)x(4/5)x(1/5)x4!/2!/2! = 24/125 ❌. Answer is half of this i.e. 12/125
d) P(AAAB) = (5/5)x(1/5)x(1/5)x(4/5)x4!/3! = 4!/3! ✅

Reference to the video: https://www.youtube.com/watch?v=cUjI2V5 ... LL&index=6
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c) 2 pairs of different colours
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August 18th DI Question of the Day





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August 18th Verbal Question of the Day





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Quote:

*jeffn sends image:*
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Mean of Any 2 students is less than 70...
Statement 1. One students weigh more than 70... means the rest should be less than 70..infact by a greater amount ... so... 1 has more than 70... hence sufficient

B. Means of all 5 students is 68...
Let all be 68... Number of students with more than 70 = 0
Let 1 by 71 and 3 be 68 and 1 be 65...
Here mean of any 2 students is less than 70 and mean of all is 68..

But we are getting 2 values... hence not sufficient..

Answer A...
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Quote:

*jeffn sends image:*
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Mean of Any 2 students is less than 70...
Statement 1. One students weigh more than 70... means the rest should be less than 70..infact by a greater amount ... so... 1 has more than 70... hence sufficient

B. Median of all 5 students is 68...
Let all be 68... Number of students with 70 or more than 70 = 0
Let 1 by 71 and 3 be 68 and 1 be 65...
Here mean of any 2 students is less than 70 and mean of all is 68..

But we are getting 2 values... hence not sufficient..

Answer A...
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10 chores = x rupees
10+ Something = 2x + something * x/10
Did y chores.. extra is y-10

1. 2x + x(y-10)/10 = 62 - NS
2. 9 chores - 20 ... NS we don’t know how much he received last sat...

Combine...
We know X = 20
Putting in Equation 1.
40 + 20(y-10)/10 = 62
Sufficient - We’ll get the value of Y

Answer C..
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Quote:

10 chores = x rupees
10+ Something = 2x + something * x/10
Did y chores.. extra is y-10

1. 2x + x(y-10)/10 = 62 - NS
2. 9 chores - 20 … NS we don’t know how much he received last sat…

Combine…
We know X = 20
Putting in Equation 1.
40 + 20(y-10)/10 = 62
Sufficient - We’ll get the value of Y

Answer C..
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Oh my bad it’s upto 10 chores

I interpreted it as if 10 chores
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Hi can anyone please tell me If that if I buy the "Pro Pack" then will I be able to reset the question pool of any one sectional test once or will I be able to reset the question pool of all the three sectional tests once?
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mayankkamal999
I am solving this question, using the Permutation method taught by Aditya but my answer in 2 of them seems to be double of the actual answer. I am not sure what mistake I am making.

There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that:
a) all the balls chosen are different colours?
b) 2 of the balls are the same colour and other 2 are of different color each?
c) 2 pairs of different colors?
d) 3 of the balls are the same colour?

Assuming we have balls of colors A,B,C,D and E
a) P(ABCD)= (5/5)x(4/5)x(3/5)x(2/5) = 24/125 ✅
b) P(AABC) = (5/5)x(1/5)x(4/5)x(3/5)x4!/2! = 144/125 ❌. Answer is half of this i.e. 72/125
c) P(AABB) = (5/5)x(1/5)x(4/5)x(1/5)x4!/2!/2! = 24/125 ❌. Answer is half of this i.e. 12/125
d) P(AAAB) = (5/5)x(1/5)x(1/5)x(4/5)x4!/3! = 4!/3! ✅
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For b) 1*1/5* 3/5*2/5* 4!/2!
For C) 1*1/5*3/5*1/5* 4!/2!*2!
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mayankkamal999
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Pratikpilot
For b) 1*1/5* 3/5*2/5* 4!/2!
For C) 1*1/5*3/5*1/5* 4!/2!*2!
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Why are we taking 3/5? Don’t we have 4 colors available since we only used one??
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mayankkamal999
Why are we taking 3/5? Don’t we have 4 colors available since we only used one??
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Since we have to choose unpaired colors, remaining of the unpaired colors to choose from is 3 from the 5
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Aboyhasnoname
10 chores = x rupees
10+ Something = 2x + something * x/10
Did y chores.. extra is y-10

1. 2x + x(y-10)/10 = 62 - NS
2. 9 chores - 20 ... NS we don’t know how much he received last sat...

Combine...
We know X = 20
Putting in Equation 1.
40 + 20(y-10)/10 = 62
Sufficient - We’ll get the value of Y

Answer C..
Show more
Does upto 10 chores mean that even on 9 chores he’ll be paid same as 10 chores??
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mayankkamal999
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Pratikpilot
Since we have to choose unpaired colors, remaining of the unpaired colors to choose from is 3 from the 5
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I’m sorry I didn’t get it. I should take out a ball of same color two times, so other 4 colors should be unused/unpaired? Can I ping you so as not to spam this chat?
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Pratikpilot
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mayankkamal999
I’m sorry I didn’t get it. I should take out a ball of same color two times, so other 4 colors should be unused/unpaired? Can I ping you so as not to spam this chat?
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Pinged you.
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