# Question 4 Exercise 6.4

Solutions of Question 4 of Exercise 6.4 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 4(i)

Three unbiased coins are tossed. What is the probability of obtaining all heads?

### Solution

First we construct a tree diagram to find out the sample space,

when the coin is tossed three times.

Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When all heads.

Let $$A=\{H H H\}$$ then $$n(A)=1$$ Hence the probability of obtaining all the heads is: $P(A)=\dfrac{n(A)}{n(S)}=\dfrac{1}{8}$.

## Question 4(ii)

Three unbiased coins are tossed. What is the probability of obtaining two heads?

### Solution

First we construct a tree diagram to find out the sample space,

when the coin is tossed three times.

Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When two heads

Let $$B=\{H T$ T.THT.TTH $\}$$ then $$n(B)=3$$ Hence the probability of gelling two heads is: $$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{3}{8}$$

## Question 4(iii)

Three unbiased coins are tossed. What is the probability of obtaining one hcad?

### Solution

First we construct a tree diagram to find out the sample space,

when the coin is tossed three times.

Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ n(S)&=2^3=8\end{align} When one head.

Let $$C=\{H H T, H T H, T H H\}$$ then $$n(C)=3$$

Hence the probability of getting one head is: $$P(C)=\dfrac{n(C)}{n(S)}-\dfrac{3}{8}$$

## Question 4(iv)

Three unbiased coins are tossed. What is the probability of obtaining at least one hcad?

### Solution

First we construct a tree diagram to find out the sample space,

when the coin is tossed three times.

Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ n(S)&=2^3=8\end{align} When at least one head.

Let \begin{align}D&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7)\\ n(D)&=7.\end{align} Hence the probability of getling at least one head is: $$P(D)=\dfrac{n(D)}{n(S)}=\dfrac{7}{8}$$

## Question 4(v)

Three unbiased coins are tossed. What is the probability of obtaining All tails?

### Solution

First we construct a tree diagram to find out the sample space,

when the coin is tossed three times.

Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ n(S)&=2^3=8\end{align} When at least two heads.

Let \begin{align}E&=\{H H H, H H T, H T H, T H H\}\\ n(E)&=4\end{align} Hence the probability of getting at least two heads is: $$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{4}{8}=\dfrac{1}{2}$$

## Question 4(vi)

Three unbiased coins are tossed. What is the probability of obtaining all tails?

### Solution

First we construct a tree diagram to find out the sample space,

when the coin is tossed three times.

Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ then n(S)&=2^3=8\end{align} When all tails.

Let \begin{align}F&=\{T T T\}\\ n(F)&=1\end{align} Hence the probability of getting all tails is: $$P(F)=\dfrac{n(F)}{n(S)}=\dfrac{1}{8}$$

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