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Temurkhon
Joined: 23 Jan 2013
Last visit: 06 Apr 2019
Posts: 408
Own Kudos:
Given Kudos: 43
Schools: Cambridge'16
02 Aug 2015, 05:22
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First option (first two balls are not red and the last is red): 6/8*5/7*2/6=5/28
Second option (first or second ball is red and the last is red): (6/8*2/7*1/6)*2=2/28
So, 5/28+2/28=7/28=1/4
A
Second option (first or second ball is red and the last is red): (6/8*2/7*1/6)*2=2/28
So, 5/28+2/28=7/28=1/4
A
13 Mar 2016, 07:02
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Could you tell me more about your way to give that formula? Thank you.
zisis
26 Feb 2021, 10:45
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Hello, please help me too:
As per my logic: P (last ball is Red) = 1 - P(First 2 balls are red)
= 1 - ( (2 / 8) * (1/7) )
= 1 - (1/28)
= 27 / 28
= 0.96
Where am I wrong ?
Thanks
As per my logic: P (last ball is Red) = 1 - P(First 2 balls are red)
= 1 - ( (2 / 8) * (1/7) )
= 1 - (1/28)
= 27 / 28
= 0.96
Where am I wrong ?
Thanks
