Question 3 Exercise 6.4
Solutions of Question 3 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 3(a)
A true or false test contains eight questions. If a student guesses the answer for each question, find the probability that $8$ answers are correct.
Solution
We have $8$ questions, each question has two options.
Therefore, The state space contains $2^8$ distinct outcomes selected without bias. Thus $$n(S)=256$$ Thus the probability for each individual outcome to occur is $$\dfrac{1}{256}$$ $8$ answers are correct.
Let $$A=\{8\}$$ Obviously only one outcome corresponds to this event,
because we can select all question to be correct by one way
i.e. $${ }^8 C_8=\dfrac{8 !}{(8-8) ! 8 !}=1$$ Therefore probability to $8$ answers are correct is: $$P(A)=\dfrac{1}{256}$$
Question 3(b)
A true or false test contains eight questions. If a student guesses the answer for each question, find the probability that $7$ answers are correct.
Solution
We have $8$ questions, each question has two options.
Therefore, The state space contains $2^8$ distinct outcomes selected without bias. $$n(S)=256$$ Thus the probability for each individual outcome to occur is $$\dfrac{1}{256}$$ $7$ answers are correct
Let $$B=\{7\}$$ then possible outcome or to select $7$ answers correct out of $8$ are: $$n(B)={ }^8 C_7=\dfrac{8 !}{(8-7) ! 7 !}=8$$ Thus the probability that $7$ answers out of $8$ are correct is: $$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{8}{256}=\dfrac{1}{32}$$
Question 3(c)
A true or false test contains eight questions. If a student guesses the answer for each question, find the probability that $6$ answers are correct.
Solution
We have $8$ questions, each question has two options.
Therefore, The state space contains $2^8$ distinct outcomes selected without bias. $$n(S)=256$$ Thus the probability for each individual outcome to occur is $\dfrac{1}{256}$
$6$ answers are correct.
Let $$C=\{6\}$$ now the ways to select $6$ out of $8$ questions are: $$n(C)={ }^8 C_6=\dfrac{8 !}{(8-6) ! 6 !}=28$$ Thus the probability that $6$ answers are correct out of $8$ is: $$P(C)=\dfrac{n(C)}{n(S)}=\dfrac{28}{256}=\dfrac{7}{64}$$
Question 3(d)
A true or false test contains eight questions. If a student guesses the answer for each question, find the probability that at least $6$ answers are correct.
Solution
We have $8$ questions, each question has two options.
Therefore, The state space contains $2^8$ distinct outcomes selected without bias. $$n(S)=256$$ Thus the probability for each individual outcome to occur is $\dfrac{1}{256}$
At least $6$ answers are correct.
Let $$D=\{6\}$$ Here $6$ question to be correct are at least maximum may be eight correct.
so the possible oulcome in this case \begin{align} n(D)&={ }^8 C_6+{ }^8 C_7+{ }^8 C_8 \\ & =28+8+1=37 . \quad \text { Thus } \\ P(D)-\dfrac{n(D)}{n(D)}&=\dfrac{37}{256}\end{align}
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