A fair coin is tossed 6 times. What is the probability of getting no

@mikemcgarry: in case 4, what about HTTHTH and HTHTTH => there are 4 possibilities :D

Very good! I stand corrected. The answer must be (A), as MonSama suggests.
Mike

Thanks for the comprehensive solution to both Mike and Monsama .. Kudos for both of you !!

All possibilities with no 2 H touching each other.

I found a similar question in the link below
a-fair-coin-is-tossed-5-times-what-is-the-probability-of-99478.html

Bunuel, is there any other way than counting method to solve this question ?

Thanks.

Dear Swaroopdev
I'm happy to respond.

My friend, a couple things to keep in mind. When Bunuel and the other experts show a solution to a problem, a relatively long and complicated solution, it's not as if we are making it complicated just for our own amusement. If there were a quick, easy, formulaic way to approach the problem, of course we would show that. In general, I would say that you can trust Bunuel and the other experts to show you the easiest, the most straightforward, and most efficient solution to any problem. Saying, "This solution looks hard. I don't like it. Can you show me an easier way?" is not the path that leads to excellence. The path that leads to excellence is all about challenging oneself to dive into what is most difficult and confusing. Assume that the path that Bunuel shows you is the optimal solution, and do your best to understand every last detail of it.

Also, there's something important thing to keep in mind about Counting and about Probability. Other branches of math, such as algebra, tend to be more formulaic and recipe based. If I give you a simple algebraic equation and ask you to solve for x, there's an easy recipe to follow. Counting and Probability are not primarily formula-based or recipe-based. Yes, there are a few formulas, but it is far from straightforward to know exactly when they can or can't be applied. Many Counting & Probability problems are about seeing the problem correct, interpreting the given information in a way that allows you to dissect the problem. Many students get into a "what should I do?" mode in problem solving, and with both Counting & Probability, it's important at the beginning of a problem to be not in the "what should I do?" mode, but in the "how do I look at this?" mode. When you are looking at the problem in the right way, what to do become obvious.

Here's a blog with a few challenging counting problems and more on the mindset you need to cultivate to be successful with these problems.
https://magoosh.com/gmat/2013/difficult- ... -problems/

Does all this make sense?
Mike

Excellent post, mikemcgarry

A big fan of your posts especially the one on different levels of "understanding". Keep them coming.

Hi mikemcgarry,

Thank you for your response. First off, i have read and learnt so much from all your posts both in verbal and quant, i really appreciate your time and effort for this too.

Also, i didn't mention this in my previous post but i didn't ask Bunuel for alternate solution just because i found other solutions as 'complicated or long or i wanted a short-cut method or solutions looks hard', i actually solved this question and arrived at the correct answer in the same way as you and others did.

The reason i asked for a alternate solution is because as we all know Bunuel and other experts too often come up with a solution which is completely different and less time consuming. That was exactly my reason behind asking for an alternate solution. Also the post was around 3 years old so was hoping if someone may have some new information on this.

Probability is one of the challenging topics for me, even though i was able to solve this problem i felt i took more time to solve it than that is necessary and i just wanted to improve on it by taking some help.

Thanks again for your advise and insights on facing problems like these.

Here is the method that I used.

1) Number of outcomes with 6T = 1
2) Number of outcomes with 5T +1H = 6!/5! = 6
3) Number of outcomes with 4T +2H = 6!/(4!*2!) = 15 [But this includes combinations where the 2Hs are together]
i.e. [HH] is placed in any of the spaces (_) in _T_T_T_T_
So, subtracting 5 from (3).
4) Number of outcomes with 3T +3H = 6!/(3!*3!) = 20 [Here again, we have to subtract combinations where the 2Hs are together]
i.e. [HH] is placed in any of the spaces (_) in {[_T_T_T_H], [_T_T_HT_], [_T_HT_T_], [_HT_T_T_]}. Realize I've removed one space (_) after the each H.
So, subtracting 4*4 = 16.
Above will take care of all combinations where 3Hs are together.

Total favourable combinations = 1+6+15-5+20-16 = 21
P = 21/64
A.

can someone explain what a fair coin and "getting no any two heads" means?

A fair coin is the one with equal probabilities of heads and tails. So, a coin for which P(tail) = P(head) = 1/2. In contrast, a biased coin would be a coin for which P(tail) ≠ P(head), for example, a coin with P(tail) = 0.6 and P(head) = 0 .4.

The question asks, about the probability of cases for which two heads does not come one after another when we toss a coin 6 times.

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