ktn11 wrote:Can someone help me on this?
First note that BOTH of these problems are outside the scope of GMAT probability, but here are some solutions to think about.
1. A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses?
There is no real formula for this one, you just need to think about the cases you want:
All Heads: There is only 1 way to get all heads
4 Heads, 1 Tail: This is the tough one because we might think that there are 5 possible placements for the one Tail, BUT to keep 3 H's in a row, we cannot put the T in the middle, so there are only 4 possibilities: HHHHT, THHHH, HTHHH, HHHTH
3 Heads, 2 Tails: Here we can just think about moving the 3 Heads as a unit so it would be the number of ways we can order a tail, a tail and the head group: HHHTT, THHHT, TTHHH
If you add all of these together, there are 1+4+3 = 8 ways to get the outcome we want. Our probability is just this over the total number of outcomes. Because there are 5 flips and 2 choices for each (Heads or Tails), the total is 2*2*2*2*2 = 2^5, so the probability is 8/32 = 1/4.
2. A fair coin is tossed 6 times. What is the probability of getting no any two heads on consecutive tosses?
This question is a MUCH more difficult problem and we need to find/assume a pattern (unless you actually know the Fibonacci based formula I'll present below - which, I might add, would be VERY strange for you to know and NEVER EVER required for ANYTHING on the GMAT). Let's look at a couple of smaller number scenarios:
Toss Once: there are 2 possible outcomes, H or T and neither result in consecutive heads, so
2 desired outcomes
Toss Twice: there are 2*2=4 possible outcomes, HH,
TT, HT, TH and 3 avoid consecutive heads, so
3 desired outcomes
Toss 3-times there are 2*2*2=8 possible outcomes, HHH, HHT, THH,
HTH, TTH, THT, HTT, TTT and 5 avoid consecutive heads, so
5 desired outcomes
Toss 4-times there are 2*2*2*2=16 possible outcomes and if you list them, 8 avoid consecutive heads, so
8 desired outcomes
...
This pattern, {2, 3, 5, 8...} is the start of the Fibonacci sequence (where a sequence in 2 means that each term is created as the sum of the 2 terms before it). That indicates that this pattern would carry...2, 3, 5, 8, 13, 21...
So for 6 tosses, the number we are looking for is 21. That will be our numerator. The denominator would be 2*2*2*2*2*2=2^6=64, so the probability is 21/64 or almost 33%.
BUT MOST IMPORTANT: This entire endeavor is WELL OUTSIDE the scope of a question you might see on the GMAT!!
Whit
The Fibonacci Side-Note:
There is a formula for the situation:
The probability that a sequence of 2 heads will not occur in a sequence of n flips We just have to calculate fib(n+2) and divide by 2^n.
The Fibonacci sequence begins with:
fib(1) = 1
fib(2) = 1
fib(3) = 2
fib(4) = 3
fib(5) = 5
fib(6) = 8
fib(7) = 13
fib(8) = 21
So for 6 flips we need fib(6+2)/2^6 = fib(8)/2^6 = 21/64. Be warned - this gets tedious for a larger number of flips and becomes a feat in complex mathematics to compute when you want to know about more than 2 consecutive flips (ie. 3 consecutive, 4 consecutive, etc...)
