The probability is 0.6 that an “unfair” coin will turn up tails on any

Question

The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?

A. 0.064
B. 0.36
C. 0.64
D. 0.784
E. 0.936

Where's my error:

Show Spoiler

P(1)+P(2)+P(3)
6/10.4/10.4/10 + 6/10.6/10.4/10 +6/10.6/10.6/10
=0.516

sriharimurthy · Accepted Answer

solve it this way:

probability that at least one of the tosses will turn up tails = 1 - probability that all will be heads

= 1 - (0.4*0.4*0.4)

= 1 - 0.064

= 0.936

where you went wrong:

probability that 1 is tails = (.6*.4*.4)+(.4*.6*.4)+(.4*.4*.6)

similarly for probability that 2 is tails.. consider all the cases..

in your solution, you have considered only one case each for both of the above..

even if you had solved it correctly, this method is longer and more complicated.. solve it as shown in my post above.. easier and quicker..

hope this helps!

cheers!

Bunuel · Answer

The problem with your solution is that 1 or 2 tails can occur in several ways:

THH can occur in 3 ways: THH, HTH and HHT (basically it's # of arrangements of 3 letters THH out of whihc 2 H's are identical, which is 3!/2!=3). So, you should have P(THH)=0.6*0.4^2*3!/2!;
Similarly TTH can also occur in 3 way: TTH, THT and HTT. So, you should have P(TTH)=0.6^2*0.4*3!/2!.
Notice that we don't have the same issue with TTT, since it can occur only in one way: TTT.

P(T\geq{1})=0.6*0.4^2*\frac{3!}{2!}+0.6^2*0.4*\frac{3!}{2!}+0.6^3=0.936 .

Answer: E.

Though this question can be solved with an easier approach:  P(T\geq{1})=1-P(H=3)=1-0.4^3=0.936  (the probability of at least one tails is 1 minus the probability of zero tails).

Answer: E.

Hope it's clear.

lagomez · Answer

I'm getting. Someone correct me if i'm wrong

First time:
probability of getting tails is .6 = .6

2nd time: 
probability of getting heads is (.4)(.6) = .24

3rd time:
probability of getting tails is (.4)(.4)(.6) = .144

Probability of flipping unfair coin and getting tails is .984

MBAhereIcome · Answer

required probability = 1 - p(3 heads)

= 1 - (\frac{2}{5})^3 = \frac{117}{125}

Van4ez · Answer

Wy we do not multiply 0.4^3by 3!/2! ? 
So as a result we will have = 1- 0.4^3*3!/2!=0.808
I do understand that this is incorrect answer, but could not explain to myself what is wrong with it.

Bunuel · Answer

Factorial correction (multiplying by x!/(y!...)) account for different ways an event can occur. For example, THH can occur in 3 ways: THH, HTH and HHT, so we are multiplying by 3!/2! = 3 (3!/2! is the number of permutations of 3 letters THH where 2 T's are the same). TTT case can occur only in one way Tails, Tails, Tails, so there is no need for factorial correction for this case. Hope it's clear. Theory on Combinations: math-combinatorics-87345.html DS questions on Combinations: search.php?search_id=tag&tag_id=31 PS questions on Combinations: search.php?search_id=tag&tag_id=52 Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

BrentGMATPrepNow · Answer

ASIDE---------------
We want P(select at least 1 tails)
When it comes to probability questions involving "at least," it's best to try using the complement. 
That is, P(Event A happening) = 1 - P(Event A  not  happening)
So, here we get: P(getting at least 1 tails) = 1 -  P( not  getting at least 1 tails) 
What does it mean to  not  get at least 1 tails? It means getting zero tails. 
So, we can write: P(getting at least 1 tails) = 1 -  P(getting zero tails) 
-------------------

P(getting zero tails)  
P(getting zero tails) = P(heads on 1st toss   AND   heads on 2nd toss   AND   heads on 3rd toss)
= P(heads on 1st toss)   x   P(heads on 2nd toss)   x   P(heads on 3rd toss)
= 0.4   x   0.4   x   0.4
=  0.064

So, P(getting at least 1 tails) = 1 -  0.064  = 0.936

Answer: E

Cheers, 
Brent

ScottTargetTestPrep · Answer

We can use the formula:

P(At least 1 tail) = 1 - P(no tails)

P(no tails) = 0.4 x 0.4 x 0.4 = 0.064

P(at least 1 tail) = 1 - 0.064 = 0.936

Answer: E

BrushMyQuant · Answer

Given that The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times. We need to find what is the probability that at least one of the tosses will turn up tails?

As the coin is tossed three times then number of cases =  2^3  = 8

P(At least one Tail) = 1 - P( 0 Tail)  = 1 - P(HHH)

P(T) = 0.6 => P(H) = 1 - 0.6 = 0.4

=> P(HHH) = 0.4*0.4*0.4 = 0.064

=>  P(At least one Tail) = 1 - P(HHH)  = 1 - 0.064 = 0.936

So,  Answer will be E 
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

https://www.youtube.com/watch?v=LRURv38kZw8

bumpbot · Answer

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