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A fair coin is flipped four times. Use a tree diagram to find the probability of observing exactly two heads. Enter your answer as a percentage, to the nearest percent.

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Mar 14,2025

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The probability of getting exactly two heads in four coin flips is approximately 38%.

Solución

We are asked to find the probability of obtaining exactly two heads when a fair coin is flipped four times. 1. **Total Outcomes:** Each coin flip has 2 possibilities, so for 4 flips, the total number of outcomes is \[ 2^4 = 16. \] 2. **Tree Diagram Approach:** A tree diagram would start at a root and branch out at each coin flip. For example, the first flip branches into \(H\) (head) and \(T\) (tail). Each of these branches again splits into \(H\) and \(T\) for the second flip, and similarly for the third and fourth flips. Although drawing the full tree diagram on paper is helpful to visualize all \(16\) outcomes, each branch represents a possible sequence of heads and tails. 3. **Counting Favorable Outcomes:** We want the outcome to have exactly 2 heads. The number of ways to choose 2 flips out of 4 to be heads is given by the combination \[ \binom{4}{2}. \] Calculating this gives: \[ \binom{4}{2} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6. \] 4. **Calculating the Probability:** The probability of exactly 2 heads is the number of favorable outcomes divided by the total number of outcomes: \[ \text{Probability} = \frac{6}{16} = 0.375. \] 5. **Expressing as a Percentage:** Converting \(0.375\) to a percentage: \[ 0.375 \times 100\% = 37.5\% \] Rounding to the nearest percent gives approximately \(38\%\). Thus, the probability of observing exactly two heads is approximately \(38\%\).

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To find the probability of getting exactly two heads when flipping a fair coin four times, we first consider the sample space. A tree diagram would show all possible outcomes (16 in total). However, we can simplify our calculations with a combination approach. The number of ways to choose 2 heads from 4 flips can be calculated using the binomial coefficient: \[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] Each sequence of flips (like HHTT, HTHT, etc.) has a probability of \( (0.5)^2 \times (0.5)^2 = 0.5^4 = \frac{1}{16} \). Now, multiply the number of favorable outcomes by the probability of each: \[ P(\text{exactly 2 heads}) = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8} \] To express this as a percentage: \[ \frac{3}{8} = 0.375 \times 100\% = 37.5\% \] Therefore, to the nearest percent, the probability of observing exactly two heads is **38%**.

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