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Sagot :
To solve this problem, we need to determine the probability of a coin landing on heads for four consecutive tosses.
1. First, recognize that the probability of a coin landing on heads in a single toss is [tex]\( \frac{1}{2} \)[/tex].
2. Since the coin tosses are independent events, the probability of getting heads in four consecutive tosses is the product of the individual probabilities of getting heads in each toss.
3. Thus, the probability of landing on heads all 4 times is calculated by multiplying [tex]\( \frac{1}{2} \)[/tex] for each of the four tosses:
[tex]\[ \left( \frac{1}{2} \right)^4 \][/tex]
4. Calculate the result of raising [tex]\( \frac{1}{2} \)[/tex] to the 4th power:
[tex]\[ \left( \frac{1}{2} \right)^4 = \frac{1}{2 \times 2 \times 2 \times 2} = \frac{1}{16} \][/tex]
5. Therefore, the probability of the coin landing on heads in all four tosses is:
[tex]\[ \frac{1}{16} \][/tex]
So, the correct answer is:
B. [tex]\( \frac{1}{16} \)[/tex]
1. First, recognize that the probability of a coin landing on heads in a single toss is [tex]\( \frac{1}{2} \)[/tex].
2. Since the coin tosses are independent events, the probability of getting heads in four consecutive tosses is the product of the individual probabilities of getting heads in each toss.
3. Thus, the probability of landing on heads all 4 times is calculated by multiplying [tex]\( \frac{1}{2} \)[/tex] for each of the four tosses:
[tex]\[ \left( \frac{1}{2} \right)^4 \][/tex]
4. Calculate the result of raising [tex]\( \frac{1}{2} \)[/tex] to the 4th power:
[tex]\[ \left( \frac{1}{2} \right)^4 = \frac{1}{2 \times 2 \times 2 \times 2} = \frac{1}{16} \][/tex]
5. Therefore, the probability of the coin landing on heads in all four tosses is:
[tex]\[ \frac{1}{16} \][/tex]
So, the correct answer is:
B. [tex]\( \frac{1}{16} \)[/tex]
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