A fair coin is tossed 4 times. What is the probability of getting at least 2 tails? Pick.
A) 1/16	B) 1/2	C) 3/16	D) 11/16	E) 3/8
Let's find the probability of the opposite event and subtract this value from 1.   The opposite event would be getting zero tails (so all heads) or 1 tail.   P(HHHH)=(12)4=116P(HHHH)=(12)4=116.   P(THHH)=4!3!∗(12)4=416P(THHH)=4!3!∗(12)4=416, we are multiplying by 4!3!4!3! since THHH scenario can occur in number of ways: THHH, HTHH , HHTH, or HHHT (notice that 4!3!4!3! basically gives number of arrangements of 4 letters THHH out of which 3 H's are identcal).   P(T≥2)=1−(116+416)=1116P(T≥2)=1−(116+416)=1116.   Answer: 11/16.
<result>D</result>
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?