Suppose we toss 8 coins simultaneously. What is the probability that at least two of them land heads?
Michael's solution:
(1) The landing of heads or tails on individual coins are independent events.
(2) The probability of a coin landing tails is $\frac12$. The probability of a coin landing heads is also $\frac12$.
(3) The probability that all eight coins land tails is $P_0=\left(\frac12\right)^8\cong 0.0039$.
(4) The probability that exactly one coin lands heads is $P_1=\left(\frac12\right)^8\cdot 8\cong 0.0313$.
(5) The probability that exactly two coins land heads is $P_2=\left(\frac12\right)^8 \cdot {8\choose2}\cong 0.1094 $
(6) The probability that at least two coins land heads is $P=P_0+P_1+P_2\cong 0.1446$.
Michael's solution is incorrect. Determine what the correct solution should look like and identify in which step Michael made a mistake.
The mistake is in step (6). The probability that at least two coins land heads is $P=1-(P_0+P_1)\cong 0.9648$
The mistake is in step (3). The probability that all eight coins land tails is $P_0=8\cdot\left(\frac12\right)^8\cong 0.0313$. The probability that at least two coins land heads is then $P=P_0+P_1+P_2\cong0.1720$.
The mistake is in step (6). The probability that at least two coins land heads in $P=1-(P_0+P_1+P_2)\cong0.8554$.
The mistake is in step (5). The probability that exactly two coins land heads is $P_2=2\cdot\left(\frac12\right)^8\cong 0.0078$. The probability that at least two coins land heads is then $P=P_0+P_1+P_2\cong 0.0430$.