If a fair coin is tossed $5$ times, what is the probability, rounded to the nearest thousandth, of getting at most $2$ tails? $\newline$ Answer:

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Grade 8

Probability of independent and dependent events

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Q. If a fair coin is tossed

5

times, what is the probability, rounded to the nearest thousandth, of getting at most

2

tails?

\newline

Answer:

Calculate Probability of $0$ Tails: To solve this problem, we need to calculate the probability of getting $0$ tails, $1$ tail, and $2$ tails in $5$ coin tosses and then sum these probabilities. $\newline$ The probability of getting a tail in one coin toss is $\frac{1}{2}$ , and the probability of getting a head is also $\frac{1}{2}$ .
Calculate Probability of $1$ Tail: First, let's calculate the probability of getting $0$ tails (which means getting $5$ heads). $\newline$ The probability of getting $5$ heads in a row is $(\frac{1}{2})^5$ .
Calculate Probability of $2$ Tails: Now, we calculate the actual probability for $0$ tails: $(\frac{1}{2})^5 = \frac{1}{32}$ .
Sum Probabilities: Next, we calculate the probability of getting exactly $1$ tail. $\newline$ This can happen in $5$ different ways (HTTTT, THTTT, TTHTT, TTTHT, TTTTH), since the tail can appear in any of the $5$ tosses. $\newline$ The probability for each of these ways is $(\frac{1}{2})^5$ .
Calculate Total Probability: Now, we calculate the actual probability for $1$ tail: $\newline$ $5 \times (\frac{1}{2})^5 = \frac{5}{32}$ .
Simplify Fraction: Next, we calculate the probability of getting exactly $2$ tails. $\newline$ This can happen in several different ways: TTTHH, THTTH, THTHT, THHTT, HTTTH, HTTHT, HTHTT, HHTTT. $\newline$ There are a total of $10$ ways to get $2$ tails in $5$ tosses (this is a combination problem, which can be calculated using ＂ $5$ choose $2$ ＂). $\newline$ The probability for each of these ways is $(\frac{1}{2})^5$ .
Round Probability: Now, we calculate the actual probability for $2$ tails: $10 \times (\frac{1}{2})^5 = \frac{10}{32}$ .
Round Probability: Now, we calculate the actual probability for $2$ tails: $10 \times (\frac{1}{2})^5 = \frac{10}{32}$ .We sum the probabilities of getting $0$ tails, $1$ tail, and $2$ tails to find the total probability of getting at most $2$ tails: Probability = Probability of $0$ tails + Probability of $1$ tail + Probability of $2$ tails Probability = $\frac{1}{32} + \frac{5}{32} + \frac{10}{32}$
Round Probability: Now, we calculate the actual probability for $2$ tails: $10 \times (1/2)^5 = 10/32$ . We sum the probabilities of getting $0$ tails, $1$ tail, and $2$ tails to find the total probability of getting at most $2$ tails: Probability = Probability of $0$ tails + Probability of $1$ tail + Probability of $2$ tails Probability = $1/32 + 5/32 + 10/32$ Now, we calculate the actual total probability: Probability = $1/32 + 5/32 + 10/32 = 16/32$
Round Probability: Now, we calculate the actual probability for $2$ tails: $10 \times (1/2)^5 = 10/32$ . We sum the probabilities of getting $0$ tails, $1$ tail, and $2$ tails to find the total probability of getting at most $2$ tails: Probability = Probability of $0$ tails + Probability of $1$ tail + Probability of $2$ tails Probability = $1/32 + 5/32 + 10/32$ Now, we calculate the actual total probability: Probability = $1/32 + 5/32 + 10/32 = 16/32$ We simplify the fraction $16/32$ to get the final probability: Probability = $1/2$
Round Probability: Now, we calculate the actual probability for $2$ tails: $\newline$ $10 \times (1/2)^5 = 10/32$ .We sum the probabilities of getting $0$ tails, $1$ tail, and $2$ tails to find the total probability of getting at most $2$ tails: $\newline$ Probability = Probability of $0$ tails + Probability of $1$ tail + Probability of $2$ tails $\newline$ Probability = $1/32 + 5/32 + 10/32$ Now, we calculate the actual total probability: $\newline$ Probability = $1/32 + 5/32 + 10/32 = 16/32$ We simplify the fraction $16/32$ to get the final probability: $\newline$ Probability = $1/2$ Finally, we round the probability to the nearest thousandth: $\newline$ Probability $\approx 0.500$