Aptitude - Probability - Discussion

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".

See, 3 coins. 3 bits. 3 bits can make numbers from 0-7 is binary:

000, 001, 010, 011, 100, 101, 110, 111.
Take 0->head, 1->tail.

So, max 2 head mean. Not more than 2 heads. We can take samples with 0 head, 1 head and 2 heads.
All except 111.
Thats it.
Total = 8.
Sample = 7.
P = 7/8.

One unbiased coin contains two sides.

So three unbiased coin contain total six sides. From the three coins when tossed together we get {H1H2T3, H1T2H3, T1H2H3}.

So probability is (3/6)= (1/2).

Okay, people have been trying to explain it to other people, but.

Why do we consider it as 2^3? I mean the total no. of outcomes, won't it be 9? (3x3) (0.0).

Isn't that how the sample is calculated?