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If a fair coin is tossed 9 times, what is the probability, rounded to the nearest thousandth, of getting at least 7 heads? Answer:

If a fair coin is tossed 99 times, what is the probability, rounded to the nearest thousandth, of getting at least 77 heads?\newlineAnswer:

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Q. If a fair coin is tossed 99 times, what is the probability, rounded to the nearest thousandth, of getting at least 77 heads?\newlineAnswer:
  1. Calculate Probabilities: To solve this problem, we need to calculate the probability of getting exactly 77 heads, exactly 88 heads, and exactly 99 heads, then add these probabilities together. The probability of getting exactly kk heads in nn tosses of a fair coin is given by the binomial probability formula:\newlineP(X=k)=(nk)⋅(pk)⋅((1−p)(n−k))P(X = k) = \binom{n}{k} \cdot (p^k) \cdot ((1-p)^{(n-k)})\newlinewhere ＂nn choose kk＂ is the binomial coefficient, pp is the probability of getting heads on a single toss (0.50.5 for a fair coin), and 8800 is the random variable representing the number of heads.
  2. Calculate 77 Heads Probability: First, we calculate the probability of getting exactly 77 heads in 99 tosses.\newlineUsing the binomial coefficient formula, we have:\newline((97)=9!(7!∗(9−7)!)=9!(7!∗2!)=(9∗8)(2∗1)=36)(9 \choose 7) = \frac{9!}{(7! * (9-7)!)} = \frac{9!}{(7! * 2!)} = \frac{(9 * 8)}{(2 * 1)} = 36\newlineNow we calculate the probability:\newline(P(X=7)=(97)∗(0.57)∗(0.59−7)=36∗(0.57)∗(0.52))P(X = 7) = (9 \choose 7) * (0.5^7) * (0.5^{9-7}) = 36 * (0.5^7) * (0.5^2)
  3. Calculate 88 Heads Probability: Next, we calculate the probability of getting exactly 88 heads in 99 tosses.\newlineUsing the binomial coefficient formula, we have:\newline((98)=9!(8!∗(9−8)!)=9!(8!∗1!)=91=9)(9 \choose 8) = \frac{9!}{(8! * (9-8)!)} = \frac{9!}{(8! * 1!)} = \frac{9}{1} = 9\newlineNow we calculate the probability:\newline(P(X=8)=(98)∗(0.58)∗(0.59−8)=9∗(0.58)∗(0.51))P(X = 8) = (9 \choose 8) * (0.5^8) * (0.5^{9-8}) = 9 * (0.5^8) * (0.5^1)
  4. Calculate 99 Heads Probability: Finally, we calculate the probability of getting exactly 99 heads in 99 tosses.\newlineUsing the binomial coefficient formula, we have:\newline((99)=9!(9!∗(9−9)!)=9!(9!∗0!)=1)(9 \choose 9) = \frac{9!}{(9! * (9-9)!)} = \frac{9!}{(9! * 0!)} = 1\newlineNow we calculate the probability:\newline(P(X=9)=(99)∗(0.59)∗(0.59−9)=1∗(0.59)∗(0.50)=(0.59))P(X = 9) = (9 \choose 9) * (0.5^9) * (0.5^{9-9}) = 1 * (0.5^9) * (0.5^0) = (0.5^9)
  5. Add Probabilities: Now we add the probabilities of getting exactly 77 heads, exactly 88 heads, and exactly 99 heads to find the total probability of getting at least 77 heads.P(X≥7)=P(X=7)+P(X=8)+P(X=9)P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9)P(X≥7)=36×(0.57)×(0.52)+9×(0.58)×(0.51)+(0.59)P(X \geq 7) = 36 \times (0.5^7) \times (0.5^2) + 9 \times (0.5^8) \times (0.5^1) + (0.5^9)
  6. Perform Calculations: We perform the calculations:\newlineP(X≥7)=36×(0.59)+9×(0.59)+(0.59)P(X \geq 7) = 36 \times (0.5^9) + 9 \times (0.5^9) + (0.5^9)\newlineP(X≥7)=(36+9+1)×(0.59)P(X \geq 7) = (36 + 9 + 1) \times (0.5^9)\newlineP(X≥7)=46×(0.59)P(X \geq 7) = 46 \times (0.5^9)\newlineP(X≥7)=46×(1/512)P(X \geq 7) = 46 \times (1/512)\newlineP(X≥7)=46/512P(X \geq 7) = 46 / 512
  7. Simplify and Round: Finally, we simplify the fraction and round to the nearest thousandth: P(X≥7)=46512≈0.0898P(X \geq 7) = \frac{46}{512} \approx 0.0898

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