Aptitude - Probability - Discussion

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.

@Belle.

A. P(All boys) = (1/2)^3 = 1/8.
B. P(All girls or all boys) = 1/8 + 1/8 = 1/4.
C. P(Exactly two boys or two girls) = 2*3C2(1/2)^2*(1/2) = 2*3*(1/8) = 3/4(here we need 2 of particular gender. So., if we calculate[2*3C2(1/2)^2] we'll get 2 of particular gender and then this should be multiplied with 1/2. Atlast we get 3/4).

D. P(At least one child of each gender) = 1 -P(all girls or all boys)= 1 - 1/4 = 3/4.
I hope this helps.

I think throwing 3 coins at a time is permutation not combination.

HTH = THH = HHT.

Reason is we did't mentioned the coins are different. So how can we say the first coin is Head second coin is Tail and third coin is Head.

If we throwing a single coin 3 times then we get combination like HHH, HHT, HTH, THH, TTT, THH, THT, TTH 8WAYS 1st throw second throw and third throw.

We can solve with combination too.

Sample space will be 2^3=8, we are taking base 2 because there are two instances; head or tails. Atmost 2 heads mean, that maximum no. Of heads include are 2, this means that the case of 0 head will also be included.

So, it will be (3c0+3c1+3c2) /8 i.e. (case with 0 head+case with 1head+case with 2 head) / sample space= 7/8.

Hi Guys,

At most two heads means "Total number of HEADs can be 0 to 2".

So we cannot include HHH, because it contains 3 heads.

But we can include TTT, because ZERO head is accepted.

Therefore, the given answers is perfectively correct.

You may ask your doubts here if you have regarding this question?

Have a nice day!

@sara.

We can get 2 heads any position out of 3 (- - -) each position has 1/2 probability. So, (1/2*1/2*1/2) + (1/2*1/2*1/2) + (1/2*1/2*1/2) =3/8.

Do the same way for 1 head (1 out of 3 positions) is (1/2*1/2*1/2) + (1/2*1/2*1/2) + (1/2*1/2*1/2) =3/8.

And last without a head is 1/8.

Add (atmost 2 to atlast 0) 1/8+3/8+3/8=7/8.

@ Vivek Rai.

A coin is tossed 3 times. We have 8 possible outcomes. In those 8 possible outcomes, we should choose the outcomes in which we have 1 head, 2 heads or no heads (3tails). But according to you if you consider the only head side of the coin then how did you get 8 in the denominator?

So the answer is 7/8.

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.

At most means maximum, that means no more than 2 heads, so the events include TTT, HTT, HHT, THH, TTH, HTH, and THT.

HHH is not right because it's more than two heads, at most means not more than two so it includes 0 heads, 1 head, 2 heads.

Guys I am confused about atleast and atmost. Can anyone give me a correct idea about that. I thought atleast 2 heads means we can have 2 heads and more than 2.

According to that we get 7/8. Then how do you get 7/8 for atmost 2 heads?