coins

[shibal]

i know it's easy, but i wanna see the different approaches (shortcuts) to the following question:

4 coins are tossed, what is the prob that at least 3 will come up heads??

oa [spoiler]5/16[/spoiler]

[pops]

Probability that all 4 shows heads: 1/2*1/2*1/2*1/2 = 1/16
Probability that 3 shows heads: 4C1*1/2*1/2*1/2*1/2 = 4/16
probability at least 3 shows heads=1/16+4/16=5/16

[tohellandback]

i would like to solve it by taking that opposite. i.e. by taking the probability of not getting at least 3 heads.

Can anyone help??

[nitya34]

is the OA ok?
I believe if we use Bionomial Theorem

we get

(4C3) [(1/2)^3][(1/2)^1]=4(1/2)^4=4/16

https://gwydir.demon.co.uk/jo/probability/info.htm

[scoobydooby]

prob of atleast 3 coins showing head:

(prob of 3 coins head, 1 coin tail)+(prob of all coins head)

nitya34, you are missing out on another possibilty- all coins showing head.
so if you add 4C4*(1/2)^4=(1/2)^4=1/16 to the 4/16 you already have, you end up with 5/16

[scoobydooby]

i would like to solve it by taking that opposite. i.e. by taking the probability of not getting at least 3 heads.

Can anyone help??

prob of getting atleast 3 heads= 1- prob of not getting atleast 3 heads.

prob of not getting atleast 3 heads= (prob of no head)+ (prob of 1 head)+ (prob of 2 head)

=4C4*(1/2)^4+4C1*(1/2)^4+4C2*(1/2)^4
=1/16+4/16+6/16
=11/16

so prob of getting atleast 3 heads=1-11/16=5/16

[Stuart@KaplanGMAT]

i know it's easy, but i wanna see the different approaches (shortcuts) to the following question:

4 coins are tossed, what is the prob that at least 3 will come up heads??

oa [spoiler]5/16[/spoiler]

Super fast way to answer coin flip questions - use Pascal's Triangle.

The n=4 row of the triangle is 1 4 6 4 1.

We want at least 3 heads, which means we add up the last two numbers of the row.. 4+1=5.. then we divide that by the sum of the row... 16.

Answer: 5/16

For more on the triangle (and other coin flip tips), take a look at https://www.beatthegmat.com/coin-flip-qu ... 17911.html