melissawlim wrote:
A fair 2-sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?
(A) 5/8
(B) 3/4
(C) 7/8
(D) 57/64
(E) 15/16
We need to determine the probability of flipping tails 2, 3, 4, or 5 times in 6 flips. Thus, we could use the following formula:
P(flipping tails 2, 3, 4, or 5 times in 6 flips) = 1 - P(selecting tails 0, 1, or 6 times)
P(tails zero times) can be denoted and calculated as:
H-H-H-H-H-H = (1/2)^6 = 1/64
P(tails 1 time) can be denoted and calculated as:
T-H-H-H-H-H = (1/2)^6 = 1/64
However, T-H-H-H-H-H can be arranged in 6!/5! = 6 ways, so the overall probability is 1/64 x 6 = 6/64.
P(tails 6 times) can be denoted and calculated as:
T-T-T-T-T-T = (1/2)^6 = 1/64
Thus:
P(flipping tails 2, 3, 4, or 5 times in 6 flips) = 1 - (1/64 + 6/64 + 1/64) = 1 - 8/64 = 1- 1/8 = 7/8.
Alternate Solution:
Let x = the number of times tails will appear. Thus, we have:
P(2 ≤ x ≤ 5) = 1 - P(x = 0) - P(x = 1) - P(x = 6)
Let’s determine P(x = 0), P(x = 1), and P(x = 6):
P(x = 0) = (½)^6 = 1/64 (Note: 0 tails means: HHHHHH)
P(x = 1) = (½)^6 x 6 = 6/64 (Note: 1 tails means: THHHHH and 5 other “1T and 5H” arrangements)
P(x = 6) = (½)^6 = 1/64 (Note: 6 tails mean: TTTTTT)
Thus:
P(2 ≤ x ≤ 5) = 1 -1/64 - 6/64 - 1/64 = 56/64 = ⅞
Answer: C