JAMB Mathematics Past Questions & Answers - Page 319

1,591.

The probability that a student passes a physics test is \(\frac{2}{3}\). If he takes three physics tests, what is the probability that he passes two of the tests?

A.

\(\frac{2}{27}\)

B.

\(\frac{3}{27}\)

C.

\(\frac{4}{27}\)

D.

\(\frac{5}{3}\)

Correct answer is C

Pass(P) = \(\frac{2}{3}\), Fail(F) = \(\frac{1}{3}\)

T = P.P.F ==> \(\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3}\) = \(\frac{4}{27}\)

1,592.

In how many ways can the letters of the word TOTALITY be arranged?

A.

6720

B.

6270

C.

6207

D.

6027

Correct answer is A

\(\frac{8!}{3!}\)

\(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2\times 1}\)

==> 8 x 7 x 6 x 5 x 4 = 6720

1,593.

Evaluate n+1Cn-2 If n =15

A.

3630

B.

3360

C.

1120

D.

560

Correct answer is D

\(\frac{n + 1 (n - 2)}{(n + 1)!}\)

\(\frac{(n + 1) + (n - 2)!(n - 2)!}{(n + 1)!}\)

\(\frac{(n + 1)(n + 1 -1)(n+1-2)(n+1-3)!}{3!(n-2)!}\)

\(\frac{(n + 1)(n)(n-1)(n-2)!}{3!(n-2)!}\)

\(\frac{(n + 1)(n)(n-1)}{3!}\)

Since n = 15

\(\frac{(15 + 1)(15)(15-1)}{3!}\)

\(\frac{16 \times 15 \times 14}{3 \times 2 \times 1}\)

= 560

1,594.

Find the standard deviation of 2,3,8,10 and 12

A.

3.9

B.

4.9

C.

5.9

D.

6.9

Correct answer is A

\(\begin{array}{c|c} x & (x - \varkappa) & (x - \varkappa)^2 \\ \hline 2 & -5 & 25 \\ \hline 3 & -4 & 16 \\ \hline 8 & 1 & 1 \\ \hline 10 & 3 & 9 \\ \hline 12 & 5 & 25 \\ \hline & & 76
\end{array}\)

S.D = \(\sqrt{\frac{(x - \varkappa)^2}{n}}\)

S.D = \(\sqrt{\frac{76}{5}}\)

S.D = 3.9

1,595.

Find the range of 4,9,6,3,2,8,10 and 11

A.

11

B.

9

C.

8

D.

4

Correct answer is B

Range = Highest Number - Lowest Number

Range = 11 - 2 = 9