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
Just counting the heads tossed,
P(3,2) = 1/8 * 1/4
P(3,1) = 1/8 * 1/2
P(3,0) = 1/8 * 1/4
P(2,1) = 3/8 * 1/2
P(2,0) = 3/8 * 1/4
P(1,0) = 3/8 * 1/4
P(A>B) = ∑P(A,B) = 1/2
That is correct. Thanks.
To determine the probability of getting more heads using coin A than coin B, we need to consider all the different outcomes for the two coins.
First, let's calculate the probability of each possible outcome for tossing coin A three times:
- 0 heads: (tails, tails, tails) = 1 combination
- 1 head: (heads, tails, tails), (tails, heads, tails), (tails, tails, heads) = 3 combinations
- 2 heads: (heads, heads, tails), (heads, tails, heads), (tails, heads, heads) = 3 combinations
- 3 heads: (heads, heads, heads) = 1 combination
Next, let's calculate the probability of each possible outcome for tossing coin B two times:
- 0 heads: (tails, tails) = 1 combination
- 1 head: (heads, tails), (tails, heads) = 2 combinations
- 2 heads: (heads, heads) = 1 combination
To figure out the probability of getting more heads with coin A than with coin B, we need to sum the probabilities of the specific outcomes where coin A has more heads.
Coin A has more heads than coin B only in two cases:
1. When coin A has 1 head (3 combinations) and coin B has 0 heads (1 combination).
2. When coin A has 2 heads (3 combinations) and coin B has 0 or 1 head (1 + 2 combinations).
So the probability of getting more heads using coin A than coin B is: (3 * 1 + 3 * (1 + 2)) / (2^3 * 2^2) = (3 + 9) / 32 = 12 / 32 = 3 / 8.
Therefore, the probability that more heads are tossed using coin A than coin B is 3/8.