JAMB Mathematics Past Questions & Answers - Page 295

1,471.

If x, y can take values from the set (1, 2, 3, 4), find the probability that the product of x and y is not greater than 6

A.

\(\frac{5}{8}\)

B.

\(\frac{5}{16}\)

C.

\(\frac{1}{2}\)

D.

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

Correct answer is A

\(\begin{vmatrix}& \hline 1 & 2 & 3 & 4\\\hline 1 & 1 & 2& 3 & 4\\2 & 2& 4 & 6 & 8\\ 3& 3& 6& 9 & 12\\4 & 4 & 8 & 12& 16\end{vmatrix}\)

p (product of x, and y \(\leq\) 6) = \(\frac{10}{16}\)

= \(\frac{5}{8}\)

1,472.

In a recent zonal championship games involving 10 teams, teams X and Y were given probabilities \(\frac{2}{5}\) and \(\frac{1}{3}\) respectively of winning the gold in the football event. What is the probability that either team will win the gold?

A.

\(\frac{2}{15}\)

B.

\(\frac{7}{15}\)

C.

\(\frac{11}{15}\)

D.

\(\frac{13}{15}\)

Correct answer is C

p(x) = \(\frac{2}{5}\) p(y) = \(\frac{1}{3}\)

p(x or y) = p(x ∪ y)

= p(x) + p(y)

= \(\frac{2}{5}\) + \(\frac{1}{3}\)

= \(\frac{11}{5}\)

1,473.

A bag contains 16 red balls and 20 blue balls only. How many white balls must be added to the bag so that the probability of randomly picking a red ball is equal to \(\frac{2}{5}\)

A.

4

B.

20

C.

24

D.

40

Correct answer is A

Number of red balls = 16,

Number of blue balls = 20

Let x represent the No of white balls to be added

∴ Total number of balls = 36 + x

2(36 + x) = 80

= 2x + 80 - 72

= 8

x = \(\frac{8}{2}\)

= 4

1,474.

Find the positive value of x if the standard deviation of the numbers 1, x + 1, 2x + 1 is 6

A.

1

B.

2

C.

3

D.

4

Correct answer is C

mean (x) = \(\frac{1 + x + 1 + 2x + 1}{3}\)

= \(\frac{3x + 3}{3}\)

= 1 + x

\(\begin{array}{c|c} X & (X -X) & (X -X)^2\\ \hline 1 & -x & x^2 \\ x + 1 & 0 & 0\\2x + 1 & x & x^2\\ \hline & & 2x^2\end{array}\)

S.D = \(\sqrt{\frac{\sum(x - 7)^2}{\sum f}}\)

= \(\sqrt{(6)}^2\)

= \(\frac{2x^2}{3}\)

= 2x2

= 18

x2 = 9

∴ x = \(\pm\) \(\sqrt{9}\)

= \(\pm\)3

1,475.

Find the variance of the numbers k, k+1, k+2,

A.

\(\frac{2}{3}\)

B.

1

C.

k + 1

D.

(k + 1)2

Correct answer is A

mean (x) = \(\frac{\sum x}{N}\)

= k + k + 1 + k + 3

= \(\frac{3k + 3}{3}\)

= k + 1

\(\begin{array}{c|c} X & (X -X) & (X -X)^2\\ \hline k & -1 & 1 \\ k + 1 & 0 & 0\\ k + 2 & 1 \\ \hline & & 2\end{array}\)

Variance (52) = \(\frac{\sum (x - x)^2}{N}\)

= \(\frac{2}{3}\)