Question

You have a biased coin with the probability of getting a head being 0.6. The probability of getting at least 1 head in 3 tosses is ____. (Rounded off to three decimal places)

Hint:

Equations of the form \(\log F = a + b\log G\) are \textbf{power laws}: \(F = e^{a}\,G^{\,b}\). For \(G=1\), \(\log G=0\), so \(F=e^{a}\) immediately.

Answer 1

Correct Answer: 0.93

Step 1 (Model).
Let \(X\) be the number of heads in 3 tosses. Then \(X\sim\text{Binomial}(n=3,p=0.6)\). We need \(P(X\ge 1)\).

Step 2 (Complement rule).
\[ P(X\ge 1)=1-P(X=0)=1-(1-p)^3=1-(0.4)^3=1-0.064=0.936. \]

Step 3 (Check by expansion).
Alternatively, \[ P(X\ge 1)=\sum_{k=1}^{3}\binom{3}{k}(0.6)^k(0.4)^{3-k} \] \[ =\binom{3}{1}(0.6)(0.4)^2+\binom{3}{2}(0.6)^2(0.4)+\binom{3}{3}(0.6)^3 \] \[ =3(0.6)(0.16)+3(0.36)(0.4)+1(0.216)=0.288+0.432+0.216=0.936. \] Rounded to three decimals \Rightarrow \(\boxed{0.936}\).