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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