Sure, let's calculate the probability of getting two or more heads when a coin is tossed 4 times.
First, let's define the probability space. When a coin is tossed 4 times, there are a total of [tex]\(2^4 = 16\)[/tex] possible outcomes. Each outcome (sequence of heads and tails) has an equal probability of occurring, since the coin is fair.
Next, we need to find the probability of getting fewer than 2 heads (i.e., 0 or 1 head) and subtract that from 1 to get the probability of getting two or more heads.
1. Probability of getting 0 heads:
- There is only 1 way to get 0 heads (all tails: TTTT).
- The probability is [tex]\( \left(\frac{1}{2}\right)^4 = \frac{1}{16} \)[/tex].
2. Probability of getting 1 head:
- There are 4 ways to get exactly 1 head (HTTT, THTT, TTHT, TTTH).
- The probability for each of these outcomes is [tex]\( \left(\frac{1}{2}\right)^4 = \frac{1}{16} \)[/tex].
- So, the total probability of getting exactly 1 head is [tex]\( 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \)[/tex].
3. Probability of getting 0 or 1 head:
- We add the probabilities of getting 0 heads and 1 head: [tex]\( \frac{1}{16} + \frac{4}{16} = \frac{5}{16} \)[/tex].
4. Probability of getting two or more heads:
- We subtract the probability of getting 0 or 1 head from 1:
[tex]\[
1 - \frac{5}{16} = \frac{16}{16} - \frac{5}{16} = \frac{11}{16}
\][/tex]
Thus, the probability of getting two or more heads when a coin is tossed 4 times is [tex]\(\frac{11}{16}\)[/tex].
So, the correct answer is:
(1) [tex]\(\frac{11}{16}\)[/tex]
