Coin A is tossed three times and coin B is tossed two times. What is the probability that more heads are tossed using coin A than coin B?
THE ANSWER IS NOT 7/16
+44 Coin \(A\)can either have more heads or less heads than Coin \(B\), but not both.
The total number of possibilities for the coin flips is \(8\cdot 4=32\).
Subtracting the number of flips that have less heads for coin \(A\)or an equal amount of heads makes \(\frac{16}{32}=\boxed{\frac{1}{2}}\).
+23246 The possible ways to toss coin B: The possible ways to toss coin A:
H H H H H H T T
H T H H T T H T
T H H T H T T H
T T T H H T T T
The probability of getting two Heads with coin B and three Heads with coin A: (1/4)(1/8) = 1/32.
The probability of getting one Head with coin B and two or three Heads with coin A: (2/4)(4/8) = 8/32.
The probability of getting no Heads with coin B and at least one Head with coin A: (1/4)(7/8) = 7/32.
Adding together these possibilities: 1/32 + 8/32 + 7/32 = 16/32 = 1/2.
