1. Good post? |

## Coin Problem

If a fair coin is tossed 4 times, whats the probability that at least 3 of the 4 tosses will come up heads.

This is a very simple problem... but i want to understand the concept behind these types of coin problems... so if someone could pls help me out with it and give a detailed explaination.

Shilpi

Ans: 5/16

2. Good post? |

## Re: Coin Problem

each time a coin is tossed, you can get either heads or tails.
so there are 2 possible outcomes for each toss.
when a coin is tossed 4 times. u will have 2*2*2*2 = 16 possible outcomes.

u want atleast 3 heads in the four tosses.

so from the 4 tosses, you can choose three tosses in which u want a head in 4C3 ways.
so these 4C3 ways will have heads. => no. of outcomes = 4C3*1 (1 because u know u have heads).

this is pretty straight forward. there can be only one possible outcome of this kind. all the four heads. or rather 4C4*1 = 1

hence total required outcomes = 4C3*1 + 4C4*1 = 5.

hence probability = 5/16

HTH.

3. Good post? |

## Re: Coin Problem

Hey thanks a lot spiderman... the explaination has cleared my doubts..

4. Good post? |
I know this topic is fairly old but isn't there a second method, that works somehow like this: P(at least) = 1 - P(event does not occur)

example: a dice is rolled twice, what is the probability that both times a number higher than 1 occurs?
number 1 occurs twice = 1/6 * 1/6 = 1/36

P(number > 1) = 1- 1/36 = 35/36

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