A fair coin is tossed 6 times. What is the probability of getting no any two heads on consecutive tosses?
Hi, there. I'm happy to help with this.
First of all, question is considerably harder and more pain-in-the-tush than what you will see on the GMAT. I don't know the source, but this seems to come from some over-achieving source that wants to give students questions much harder than the test.
So this is a probability question that is best solved with counting. You may find this blog germane:http://magoosh.com/gmat/2012/gmat-quant-how-to-count/
Probability = (# of desired cases)/(total # of possible cases)
The denominator is very easy --- two possibilities for each toss, six tosses, so 2^6 = 64. That's the denominator.
For the numerator, we have to sort through cases:Case One: six tails
For this case, obviously you can't have two heads in a row. There's only one way this can happen: TTTTTTONECase Two: five tails, one head
Again, it's impossible to have two heads in a row, because there's only one. There are six ways this could happen --- the H could occupy any of the six positions (HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, and TTTTTH)SIXCase Three: four tails, two heads
This is the tricky case. There are 6C2 = 15 places that the two H's could land, but five of those (HHTTTT, THHTTT, TTHHTT, TTTHHT, and TTTTHH) involve the pair of H's together, which is forbidden. Excluding those five forbidden cases, we are left with 15 - 5 = 10 possibilities here. TENCase Four: three tails, three heads
Now, things are starting to get crowded. We have to space our three H's out, with T's between them, so none of the H's touch. That leaves only two possibilities: HTHTHT and THTHTH. That's it: any other configuration would have two H's next to each other, which is forbidden. TWO
Add those up: 1 + 6 + 10 + 2 = 19. That's our numerator.
Probability = 19/64
Answer = C
Does all that make sense? Please let me know if you have any further questions.
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