altairahmad wrote:
Same question please :
Why isn't probability of At Lease 1 head AND 1 tail = Probability of At Least 1 head x Probability of At Last 1 Tail ? i.e (31/32)*(31/32).
Bunuel GMATPrepNow chetan2uWill appreciate a conceptual explanation and any links to relevant resources.
Hi,
Why it is not \(\frac{31}{32}*\frac{31}{32}\)?
The EVENT of flipping the coin is only one. At least one head and at least one tail are NOT talking of different events..So, when will it be \(\frac{31}{32}*\frac{31}{32}\)?
If the question was that he flips coins 5 times twice. In the first he gets at least one head and in other he gets at least one tail.What is the question here?
Quote:
If a child flips a coin five times in a row, what is the probability that she will receive at least one head and one tail?
So we are talking of SAME 5 flips where there is at least one head and at least one tail is there.TWO ways to do it..
(I) Find when this does not happen and subtract from total ways..When will we NOT get at least one head and at least one tail? -
when all are head or all are tail. -- 2 waysTotal ways \(2*2*2*2*2 = 32\) ---
since 2 outcomes for each flipProbability that all are head or all are tail = \(\frac{2}{32}=\frac{1}{16}\), so probability that at least one head and at least one tail = \(1-\frac{1}{16}=\frac{15}{16}\)
(II) Find ways in which we have at least one head and at least one tail1 head and 4 tails -- 5C4 = 5 ways
2 heads and 3 tails -- 5C3 = 10 ways
3 heads and 2 tails -- 5C2 = 10 ways
4 heads and 31 tail -- 5C1 = 5 ways
Total possibilities = 5+10+10+5=30
Total ways = 2*2*2*2*2=32Probability = \(\frac{30}{32}=\frac{15}{16}\)
C