MathRevolution wrote:
If you are flipping the coin 4 times, what is the probability of landing on head two times?
A. \(\frac{3}{5}\)
B. \(\frac{4}{5}\)
C. \(\frac{3}{7}\)
D. \(\frac{4}{7}\)
E. \(\frac{3}{8}\)
Two results for each of four coin flips. When ways to perform tasks in series, we multiply. So that is 2×2×2×2 results in total. That is 2[square]4 or 16.
For the favourable case we need to count the ways to get 2 heads and 2 tails. The count of permutations of two pairs of symbols is: 4!/2!2!=6. This is easily confirmed by just counting.
|HHTT,HTHT,HTTH,THHT,THTH,TTHH}=6
Thus the probability is: 6/16, or:
0.375