Aptitude - Probability - Discussion

How many times a man can tossing a coin so that the probability of atleast one head is more than 80%?

SOURAV

If NOW will be together all the time all the time then take "NOW" as one letter. Lucknow will be considered as 5 letters. Then you can get the answer easily.

You ask most two head but why you considered zero head?

Is there any short way to get (s) ?. This way takes much time.

P(2 heads)= Total favourable events divided by total events. Favourable are only 2 so then why would we take 7 on the numerator ?

For example, we want at least 2 heads from 3 tosses of coin.

Biased coin has two output, therefore for n tosses of are output is r^n i.e. 2^3 = 8 possible arrangements.

In probability.

"At least" 2 heads = more than 2 head includes 2 heads e.g. HHH, HH?, ?H?, ?HH.

"Less than" 2 heads = neither HHH nor even HH?, ?HH, H?H.

"More than" 2 heads = only HHH.

"At most" 2 heads= {TTT, TTH, THT, HTT, THH, HTH, HHT}.

Therefore, for at most 2 heads we have 7 arrangements.

From total 8 possible outcomes the probability is 7/8.

Simple and smart solution:

Find probability of getting 3 head. then subtract it from 1.

2C1*2C1*2C1 = 8.

Now 1/8 is probability,

So 1 - 1/8 = 7/8.

Can you explain why were taken 7/8 as output why were taken 7?

Because of,
Sample space is n(S)=8 this is we are tossing the coins,
n(E)=7.

So probability is n(E)/n(S) = 7/8.

If the question would be to find out the the probability of getting at least one head then it must be 7/8 in this case also. Thanks :"Indiabix".