# Discrete Mathematics Questions and Answers – Counting – Terms in Binomial Expansion

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

Explanation: Here, n = 8, p = 0.6, q = 0.4. Suppose X = number of hits x0 = 0 number of hits, x1 = 1 hit, x2 = 2 hits, and so on.
So, (X) = P(x5) + P(x6) + P(x7) + P(x8) = 8C5(0.6)5(0.4)3 + 8C5(0.6)6(0.4)2 + 8C7(0.6)7(0.4)1 + 8C8(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

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 = 15C0(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

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) $$\frac{3}{97}$$

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, $$\frac{5}{6}$$) = 2C2 * ($$\frac{4}{6}$$)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

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) = 20C11(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*(x5)
b) $$\frac{1}{5}*(\frac{x}{4})$$
c) 10*($$\frac{x}{3}$$)
d) 6*(x3)

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) 3y2 + 2y5
b) 7(y4 + y2 + y)
c) 8(y3 + y1)
d) y + y2 + y3

Explanation: By using Binomial theorem,the expression (y+1)4 – (y-1)4 can be expanded as = (y+1)4 = 4C0y4 + 4C1y3 + 4C2y2 + 4C3y1 + 4C4y0 and (y-1)4 = 4C0y44C1y3 + 4C2y24C3y1 + 4C4y0. Now, (y+1)4 – (y-1)4 = (4C0y4 + 4C1y3 + 4C2y2 + 4C3y1 + 4C4y0) – (4C0y44C1y3 + 4C2y24C3y1 + 4C4y0) = 2(4C1y3 + 4C3y1) = 8(y3 + y1).

8. Find the coefficient of x7 in (x+4)9.
a) 523001
b) 428700
c) 327640
d) 129024

Explanation: It is known that (r+1)th term, in the binomial expansion of (a+b)n is given by, Tr+1 = nCran-rbr. Assuming that x7 occurs in the (r+1)th term of the expansion (x+4)9, we obtain Tr+1 = 129024x4.

9. Determine the 7th term in the expansion of (x-2y)12.
a) 6128y7
b) 59136y6
c) 52632x6
d) 39861y5

Explanation: By assuming that x7 occurs in the (r+1)th term of the expansion (x-2y)12, we obtain Tr+1 = nCran-rbr = 12C6 x6 (2y)6 = 59136y6.

10. What is the middle term in the expansion of (x/2 + 6y)8?
a) 45360x4
b) 34210x3
c) 1207x4
d) 3250x5

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 = 5th term.
Now, assuming that x5 occurs in the (r+1)th term of the expansion (x/2+6y)8, we obtain Tr+1 =nCrxn-ryr = 8C4(x/2)4(6y)4 = 45360x4.

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