John flips a coin 4 times. What is the probability that he gets heads

The "1-x" short cut above is best here, but you can verify the long way.

There are 16 possible combinations here. They break down as follows:

1H, 3T: 4 ways (heads first, second, third, or fourth)
2H, 2T: 6 ways (heads in 1&2, 1&3, 1&4, 2&3, 2&4, or 3&4)
3H, 1T: 4 ways (as in the first one, but this time tails first, second, third, or fourth)
4H, 0T: 1 way (HHHH)
0H, 1T: 1 way (TTTT)

There are your 16 ways. 15/16 involve heads. Aren't you glad we have the 1-x trick?

If we assume that each individual coin is equally likely to come up heads or tails, then each of the above 16 outcomes to 4 flips is equally likely. Each occurs a fraction one out of 16 times, or each has a probability of 1/16.

Alternatively, we could argue that the 1st coin has probability 1/2 to come up heads or tails, the 2nd coin has probability 1/2 to come up heads or tails, and so on for the 3rd and 4th coins, so that the probability for any one particular sequence of heads and tails is just (1/2)x(1/2)x(1/2)x(1/2)=(1/16).

Now lets ask: what is the probability that in 4 flips, one gets N heads, where N=0, 1, 2, 3, or 4. We can get this just by counting the number of outcomes above which have the desired number of heads, and dividing by the total number of possible outcomes, 16.



N # outcomes with N heads probability to get N heads

0 1 1/16 = 0.0625

1 4 4/16 = 1/4 = 0.25

2 6 6/16 = 3/8 = 0.375

3 4 4/16 = 1/4 = 0.25

4 1 1/16 = 0.0625

John flips a coin 4 times. What is the probability that he gets heads on at least one of the four flips?

(A) 1/16
(B) 1/4
(C) 1/2
(D) 3/4
(E) 15/16

John flips a coin 4 times. What is the probability that he gets heads on at least one of the four flips?

(A) 1/16
(B) 1/4
(C) 1/2
(D) 3/4
(E) 15/16