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What is the probability of tossing 21 or more heads or more out of 30 tosses? Answer to 2 decimal places as a percent.Ex: if your answer is 4.545% answer 4.55 Use Stapplet. square

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What is the probability of tossing 21 or more heads or more out of 30 tosses?
Answer to 2 decimal places as a percent.Ex: if your answer is 4.545%  answer
4.55
Use Stapplet.
square

What is the probability of tossing 21 or more heads or more out of 30 tosses? Answer to 2 decimal places as a percent.Ex: if your answer is 4.545% answer 4.55 Use Stapplet. square

Solución

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Robertomaestro · Tutor durante 5 años
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To find the probability of getting 21 or more heads out of 30 tosses, we can use the binomial probability formula. The binomial probability formula is given by:<br /><br />$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$<br /><br />where:<br />- $n$ is the number of trials (in this case, 30 tosses)<br />- $k$ is the number of successes (in this case, 21 or more heads)<br />- $p$ is the probability of success on a single trial (in this case, 0.5 for a fair coin)<br /><br />We need to calculate the probability of getting 21, 22, 23,..., up to 30 heads. This can be done by summing the individual probabilities for each value of $k$ from 21 to 30.<br /><br />$P(X \geq 21) = P(X = 21) + P(X = 22) + P(X = 23) + \ldots + P(X = 30)$<br /><br />Using the binomial probability formula, we can calculate each term individually and then sum them up.<br /><br />$P(X = 21) = \binom{30}{21} (0.5)^{21} (0.5)^{9}$<br />$P(X = 22) = \binom{30}{22} (0.5)^{22} (0.5)^{8}$<br />$P(X = 23) = \binom{30}{23} (0.5)^{23} (0.5)^{7}$<br />$\vdots$<br />$P(X = 30) = \binom{30}{30} (0.5)^{30} (0.5)^{0}$<br /><br />Now, we can calculate each term using the binomial coefficient and plug in the values of $n$, $k$, and $p$.<br /><br />$P(X = 21) = \frac{30!}{21! \cdot 9!} \cdot (0.5)^{21} \cdot (0.5)^{9}$<br />$P(X = 22) = \frac{30!}{22! \cdot 8!} \cdot (0.5)^{22} \cdot (0.5)^{8}$<br />$P(X = 23) = \frac{30!}{23! \cdot 7!} \cdot (0.5)^{23} \cdot (0.5)^{7}$<br />$\vdots$<br />$P(X = 30) = \frac{30!}{30! \cdot 0!} \cdot (0.5)^{30} \cdot (0.5)^{0}$<br /><br />After calculating each term, we can sum them up to get the final probability.<br /><br />$P(X \geq 21) = P(X = 21) + P(X = 22) + P(X = 23) + \ldots + P(X = 30)$<br /><br />Finally, we can convert the probability to a percentage by multiplying it by 100.<br /><br />$P(X \geq 21) \times 100\%$<br /><br />The final answer will be the probability of getting 21 or more heads out of 30 tosses, expressed as a percentage.
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