\frac{5}{33}
\frac{13}{66}
\frac{8}{53}
\frac{19}{66}
Correct answer is D
P(\text{two same color balls}) = P(\text{2 red}) + P(\text{2 white}) + P(\text{2 black})
P(\text{2 red}) = \frac{3}{12} \times \frac{2}{11} = \frac{1}{22}
P(\text{2 white}) = \frac{4}{12} \times \frac{3}{11} = \frac{1}{11}
P(\text{2 black}) = \frac{5}{12} \times \frac{4}{11} = \frac{5}{33}
P(\text{2 same color balls}) = \frac{1}{22} + \frac{1}{11} + \frac{5}{33} = \frac{19}{66}
Eight football clubs are to play in a league on home and away basis. How many matches are possible?
14
28
56
128
Correct answer is C
Number of matches possible = 2 \times ^{8}C_{2} (Home and away repitition of the matches)
= 2 \times \frac{8!}{(8 - 2)! 2!}
= 2 \times 28
= 56
If the mean of -1, 0, 9, 3, k, 5 is 2, where k is a constant, find the median of the set of numbers.
\frac{3}{2}
0
\frac{7}{2}
6
Correct answer is A
\frac{-1 + 0 + 9 + 3 + k + 5}{6} = 2 \implies 16 + k = 12
k = -4
Arranging -1, 0, 9, 3, -4, 5 in order: -4, -1, 0, 3, 5, 9
Median = \frac{0 + 3}{2} = \frac{3}{2}
Evaluate \begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix}.
(13, 11)
(11, 13)
\begin{pmatrix} 13 \\ 11 \end{pmatrix}
\begin{pmatrix} 11 \\ 13 \end{pmatrix}
Correct answer is C
\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix}
= \begin{pmatrix} 2 \times 2 + 3 \times 3 \\ 4 \times 2 + 1 \times 3 \end{pmatrix}
= \begin{pmatrix} 13 \\ 11 \end{pmatrix}