Respuesta :
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Explain the concept
Tossing a coin, Sample space (S) = {H,T}
Total outcome n(S) = 2
Probability =
[tex]Probability=\frac{number\text{ of required outcomes}}{number\text{ of total outcomes}}[/tex]Conditional probablity: P(A|B) =
[tex]\frac{P(A\cap B)}{P(B)}=\frac{n(A\cap B)}{n(B)}[/tex]STEP 2: Calculate the probability
A fair coin tossed three times will have the sample space (S) given as:
[tex]\begin{gathered} S=\lbrace(HHH),(HHT),(HTH),(THH),(TTH),(THT),(HTT),(TTT)\rbrace \\ n(S)=8 \end{gathered}[/tex]Let, event for getting at least two tails is A
[tex]\begin{gathered} A=\lbrace(TTT,TTH,THT,HTT)\rbrace \\ n(A)=4 \end{gathered}[/tex]Let, event for getting at least one tail is B
[tex]\begin{gathered} B=\lbrace(TTT),(TTH),(THT),(HTT),(HHT),(HTH),(THH) \\ n(B)=7 \end{gathered}[/tex]Now,
[tex]\begin{gathered} (A\cap B)=\lbrace(TTT),(TTH),(THT),(HTT)\rbrace \\ n(A\cap B)=4 \end{gathered}[/tex]The answer is therefore calculated as:
[tex]P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}=\frac{n(A\cap B)}{n(B)}=\frac{4}{7}[/tex]Hence, the answer is 4/7
