I toss a coin 4 times. Find the chance of getting
a) 2 heads
b) more heads than tails
To find the chance of getting a specific outcome when tossing a coin multiple times, we need to use probability.
a) To find the chance of getting 2 heads when tossing a coin 4 times, we can use the binomial probability formula. The formula is: P(x) = C(n, x) * p^x * (1-p)^(n-x), where P(x) is the probability of getting x number of successful outcomes (in this case, heads), n is the total number of trials (coin tosses), C(n, x) is the number of combinations of n items taken x at a time, p is the probability of success (getting a head), and (1-p) is the probability of failure (getting a tail).
In this case, we have n = 4 (4 coin tosses) and x = 2 (2 heads). The probability of getting a head in a fair coin toss is 0.5, so p = 0.5. Plugging in these values, we get:
P(2 heads) = C(4, 2) * (0.5)^2 * (1-0.5)^(4-2)
= 6 * 0.25 * 0.25
= 0.375
Therefore, the chance of getting exactly 2 heads when tossing a coin 4 times is 0.375 or 37.5%.
b) To find the chance of getting more heads than tails, we need to consider all possible outcomes that fulfill this condition out of the total possible outcomes.
There are 2^4 = 16 possible outcomes when tossing a coin 4 times (heads or tails for each toss). We need to identify the outcomes where the number of heads is greater than the number of tails.
The possible outcomes with more heads than tails are:
- HHTT
- HTHH
- HTTH
- THHH
- TTHH
- TTTH
- TTHT
- THTT
There are a total of 8 outcomes that satisfy this condition out of the 16 possible outcomes.
Therefore, the chance of getting more heads than tails when tossing a coin 4 times is 8/16 or 0.5, which is equivalent to 50%.