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
View Answer

Answer: c
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.
advertisement

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

Answer: c
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
View Answer

Answer: c
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

Answer: a
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
View Answer

Answer: c
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.
advertisement

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)
View Answer

Answer: c
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
View Answer

Answer: c
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
View Answer

Answer: d
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
View Answer

Answer: b
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.
advertisement

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

Answer: a
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.

Sanfoundry Global Education & Learning Series – Discrete Mathematics.

To practice all areas of Discrete Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn