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(question from someone terrible in math)

Does it make sense to apply this formula on below question from OG11? If so, how can I do it? I have tried with no success.

From OG11 page 241 ex 173

THE PROBABILITY IS 1/2 THAT A CERTAIN COIN WILL TURN UP HEADS ON ANY GIVEN TOSS. IF THE COIN IS TO BE TOSSED THREE TIMES, WHAT IS THE PROBABILITY THAT ON AT LEAST ONE OF THE TOSSES THE COIN WILL TURN UP TAILS?

A) 1/8
B) 1/2
C) 3/4
D) 7/8
E) 15/16

I think the strategy outlined is great. However, for coin problems or any problems where you have multiple trials and the probabilities stay the same I like to use bernouli's formula:
(nCr)p^r x q^n-r

n = number of trials
r = number of specific events you wish to obtain
p = probability that the event will occur
q = probability that the event will not occur

I know it looks scary but it's easy to use.
Here are the steps to using it on this problem.

Question:
Solution:
At least one tail means 1 tail or 2 tails or 3tails. OR means we have to add probabilities.
P(1 tail) + P(2tails) + P(3 Tails)
We can solve this but have to find 3 probabilities. Not fun.


When you see the word at least immediately think of the reverse situation as it saves time.
In reverse what's the condition where we don't have at least 1 tail. That would be HHH or all heads.
We find the probability of the event not occurring( i.e all heads) and minus this from 1.

P(at least one tail) = 1 - (event not occuring)
To use bernouli we write down:
sucesses: P(H) = 1/2 We want 3 heads
failures: P(T) = 1/2 We want 0 tails
number of trials = 3 flips
We want 3 heads and 0 tails
Our successes in this case would be getting a head. So out of 3 flips we want to choose 3 heads or 3C3.
Here's what the formula looks like:
(1/2)³ x (1/2)^0 x 3C3
(1/2)³ x (1/2)^0 = 1/8
3C3 = 3!/3!0!= 1
So the probability of getting 3 heads and zero tails after 3 trials is 1/8
Therefore,
P(at least one tail) = 1 - 1/8 = 7/8

_________________
700+ your target then check out my 800 gmat blog here:
http://800gmatblog.zoxic.com

nice. many tks!

Using nCk/2^n

for one HEADS = 5C1/2^5 = 5/32

So for at least two:

1-(5/32) = 27/32

What am I missing here ?

_________________
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

The probability of getting 1 head should be.
(1/2)^1 x (1/2) ^4 x 5C1
1/32 x (5!/4!1!) = 5/32
Thus,
the probability of getting at least 2 heads should be
P(at least 2 heads) 1 - 5/32 = 27/32
Logitech you are quite correct. Perhaps Stuart Kovinsky from Kaplan can comment on this.

_________________
700+ your target then check out my 800 gmat blog here:
http://800gmatblog.zoxic.com

Thank you Stuart!

_________________
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

I love your formula. Thank you for inventing such a convenient formula. However, I don't know why I need to pick the third numbe rof your first formula regarding to the questions of 2 heads in a 3 flips. and so as the second and third questions. Can you please explain to me why you pick the third number but not the second number for 2 heads in 3 flips?

Thank you
mm

what is the strategy to fill the numbers in the pascal triangle..
like, if I want to form a row for 7 coin flips, how to I do that?

btw, Great info... thanks a ton

Icredible tool!!
Thanks