Binomial Expansion Terms Questions and Answers

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.

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.

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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 = ⁴C₀y⁴ – ⁴C₁y³ + ⁴C₂y² – ⁴C₃y¹ + ⁴C₄y⁰. Now, (y+1)⁴ – (y-1)⁴ = (⁴C₀y⁴ + ⁴C₁y³ + ⁴C₂y² + ⁴C₃y¹ + ⁴C₄y⁰) – (⁴C₀y⁴ – ⁴C₁y³ + ⁴C₂y² – ⁴C₃y¹ + 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, T_r+1 = ⁿC_ra^n-rb^r. Assuming that x⁷ occurs in the (r+1)^th term of the expansion (x+4)⁹, we obtain T_r+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 T_r+1 = ⁿC_ra^n-rb^r = 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
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 = 5^th term.
Now, assuming that x⁵ occurs in the (r+1)^th term of the expansion (x/2+6y)⁸, we obtain T_r+1 =ⁿC_rx^n-ry^r = ⁸C₄(x/2)⁴(6y)⁴ = 45360x⁴.

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