Question

You toss a fair coin three times. Given that you have observed at least one tails, what is the probability that you observe at least two tails?3/74/71/21

Answer 1

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 =

`Probability=\frac{number\text{ of required outcomes}}{number\text{ of total outcomes}}`

Conditional probablity: P(A|B) =

`(P(A\cap B))/(P(B))=(n(A\cap B))/(n(B))`

STEP 2: Calculate the probability

A fair coin tossed three times will have the sample space (S) given as:

`\begin{gathered} S=\lbrace(HHH),(HHT),(HTH),(THH),(TTH),(THT),(HTT),(TTT)\rbrace \\ n(S)=8 \end{gathered}`

Let, event for getting at least two tails is A

`\begin{gathered} A=\lbrace(TTT,TTH,THT,HTT)\rbrace \\ n(A)=4 \end{gathered}`

Let, event for getting at least one tail is B

`\begin{gathered} B=\lbrace(TTT),(TTH),(THT),(HTT),(HHT),(HTH),(THH) \\ n(B)=7 \end{gathered}`

Now,

`\begin{gathered} (A\cap B)=\lbrace(TTT),(TTH),(THT),(HTT)\rbrace \\ n(A\cap B)=4 \end{gathered}`

The answer is therefore calculated as:

`P((A)/(B))=(P(A\cap B))/(P(B))=(n(A\cap B))/(n(B))=(4)/(7)`

Hence, the answer is 4/7