This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Counting – Terms in Binomial Expansion”.

1. In a blindfolded game, a boy can hit the target 8 times out of 12. If he fired 8 shots, find out the probability of more than 4 hits?

a) 2.530

b) 0.1369

c) 0.5938

d) 3.998

View Answer

Explanation: Here, n = 8, p = 0.6, q = 0.4. Suppose X = number of hits x

_{0}= 0 number of hits, x

_{1}= 1 hit, x

_{2}= 2 hits, and so on.

So, (X) = P(x

_{5}) + P(x

_{6}) + P(x

_{7}) + P(x

_{8}) =

^{8}C

_{5}(0.6)

^{5}(0.4)

^{3}+

^{8}C

_{5}(0.6)

^{6}(0.4)

^{2}+

^{8}C

_{7}(0.6)

^{7}(0.4)

^{1}+

^{8}C

_{8}(0.6)

^{8}(0.4)

^{0}= 0.5938.

2. A fair coin is tossed 15 times. Determine the probability in which no heads turned up.

a) 2.549 * 10^{-3}

b) 0.976

c) 3.051 * 10^{-5}

d) 5.471

View Answer

Explanation: According to the null hypothesis it is a fair coin and so in that case the probability of flipping at least 59% tails is =

^{15}C

_{0}(0.5)

^{15}= 3.051 * 10

^{-5}.

3. When a programmer compiles her code there is a 95% chance of finding a bug every time. It takes three hours to rewrite her code when she finds out a bug. Determine the probability such that she will finish her coding by the end of her workday. (Assume, a workday is 7 hours)

a) 0.065

b) 0.344

c) 0.2

d) 3.13

View Answer

Explanation: A success is a bug-free compilation, and a failure is the finding out of a bug. The programmer has 0, 1, 2, or 3 failures and so her probability of finishing the program is : Pr(X=0) + Pr(X=1) + Pr(X=2) + Pr(X=3) = (0.95)

^{0}(0.05) + (0.95)

^{0}(0.05) + (0.95)

^{0}(0.05) + (0.95)

^{0}(0.05) = 0.2.

4. Determine the probability when a die is thrown 2 times such that there are no fours and no fives occur?

a) \(\frac{4}{9}\)

b) \(\frac{56}{89}\)

c) \(\frac{13}{46}\)

d) \(\frac{3}{97}\)

View Answer

Explanation: In this experiment, throwing a die anything other than a 4 is a success and rolling a 4 is failure. Since there are two trials, the required probability is

b(2; 2, \(\frac{5}{6}\)) =

^{2}C

_{2}* (\(\frac{4}{6}\))

^{2}* (\(\frac{2}{6}\))

^{0}= \(\frac{4}{9}\).

5. In earlier days, there was a chance to make a telephone call would be of 0.6. Determine the probability when it could make 11 successes in 20 attempts of phone call.

a) 0.2783

b) 0.2013

c) 0.1597

d) 3.8561

View Answer

Explanation: Probability of success p=0.6 and q=0.4. X=success in making a telephone call. Hence, the probability of 11 successes in 20 attempts = P(X=11) =

^{20}C

_{11}(0.6)

^{11}(0.4)

^{20}– 11 = 0.1597.

6. By the expression \(\left(\frac{x}{3} + \frac{1}{x}\right)^5\), evaluate the middle term in the expression.

a) 10*(x^{5})

b) \(\frac{1}{5}*(\frac{x}{4})\)

c) 10*(\(\frac{x}{3}\))

d) 6*(x^{3})

View Answer

Explanation: By using Binomial theorem,the expression \(\left(\frac{x}{3} + \frac{1}{x}\right)^5\) can be expanded as \(\left(\frac{x}{3} + \frac{1}{x}\right)^5 = ^5C_0(\frac{x}{3})^5 + ^5C_1(\frac{x}{3})^4(\frac{1}{x})^1 + ^5C_2(\frac{x}{3})^3(\frac{1}{x})^2\)

\(+ ^5C_3(\frac{x}{3})^2(\frac{1}{x})^3 + ^5C_4(\frac{x}{3})^1(\frac{1}{x})^4 \) = \((\frac{x}{3})^5 + 5.(\frac{x}{3}) + 10.(\frac{x}{3}) + 10.(\frac{1}{3x}) + 5(\frac{1}{3x^3})\). Hence, the middle term is 10*(\(\frac{x}{3}\)).

7. Evaluate the expression (y+1)^{4} – (y-1)^{4}.

a) 3y^{2} + 2y^{5}

b) 7(y^{4} + y^{2} + y)

c) 8(y^{3} + y^{1})

d) y + y^{2} + y^{3}

View Answer

Explanation: By using Binomial theorem,the expression (y+1)

^{4}– (y-1)

^{4}can be expanded as = (y+1)

^{4}=

^{4}C

_{0}y

^{4}+

^{4}C

_{1}y

^{3}+

^{4}C

_{2}y

^{2}+

^{4}C

_{3}y

^{1}+

^{4}C

_{4}y

^{0}and (y-1)

^{4}=

^{4}C

_{0}y

^{4}–

^{4}C

_{1}y

^{3}+

^{4}C

_{2}y

^{2}–

^{4}C

_{3}y

^{1}+

^{4}C

_{4}y

^{0}. Now, (y+1)

^{4}– (y-1)

^{4}= (

^{4}C

_{0}y

^{4}+

^{4}C

_{1}y

^{3}+

^{4}C

_{2}y

^{2}+

^{4}C

_{3}y

^{1}+

^{4}C

_{4}y

^{0}) – (

^{4}C

_{0}y

^{4}–

^{4}C

_{1}y

^{3}+

^{4}C

_{2}y

^{2}–

^{4}C

_{3}y

^{1}+

^{4}C

_{4}y

^{0}) = 2(

^{4}C

_{1}y

^{3}+

^{4}C

_{3}y

^{1}) = 8(y

^{3}+ y

^{1}).

8. Find the coefficient of x^{7} in (x+4)^{9}.

a) 523001

b) 428700

c) 327640

d) 129024

View Answer

Explanation: It is known that (r+1)

^{th}term, in the binomial expansion of (a+b)

^{n}is given by, T

_{r+1}=

^{n}C

_{r}a

^{n-r}b

^{r}. Assuming that x

^{7}occurs in the (r+1)

^{th}term of the expansion (x+4)

^{9}, we obtain T

_{r+1}= 129024x

^{4}.

9. Determine the 7th term in the expansion of (x-2y)^{12}.

a) 6128y^{7}

b) 59136y^{6}

c) 52632x^{6}

d) 39861y^{5}

View Answer

Explanation: By assuming that x

^{7}occurs in the (r+1)

^{th}term of the expansion (x-2y)

^{12}, we obtain T

_{r+1}=

^{n}C

_{r}a

^{n-r}b

^{r}=

^{12}C

_{6}x

^{6}(2y)

^{6}= 59136y

^{6}.

10. What is the middle term in the expansion of (x/2 + 6y)^{8}?

a) 45360x^{4}

b) 34210x^{3}

c) 1207x^{4}

d) 3250x^{5}

View Answer

Explanation: We know that in the expansion of (x+y)

^{n}, if n is even then the middle term is (n/2 + 1)

^{th}term. Hence, the middle term in the expansion of (x/2 + 6y)

^{8}is (8/2+1)

^{th}= 5

^{th}term.

Now, assuming that x

^{5}occurs in the (r+1)

^{th}term of the expansion (x/2+6y)

^{8}, we obtain T

_{r+1}=

^{n}C

_{r}x

^{n-r}y

^{r}=

^{8}C

_{4}(x/2)

^{4}(6y)

^{4}= 45360x

^{4}.

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