Exams Joint Entrance Examination Mathematics Probability Question 13 Question 13 of 417 Medium MCQ
Let the mean and variance of 7 observations 2 , 4 , 10 , x , 12 , 14 , y , x > y 2,4,10, x, 12,14, y, x>y 2 , 4 , 10 , x , 12 , 14 , y , x > y , be 8 and 16 respectively. Two numbers are chosen from { 1 , 2 , 3 , x − 4 , y , 5 } \{1,2,3, x-4, y, 5\} { 1 , 2 , 3 , x − 4 , y , 5 } one after another without replacement, then the probability, that the smaller number among ...
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= 8 2 + 4 + 10 + x + y + 12 + 14 7 = 8 ⇒ x + y = 14 = 16 16 = 2 2 + 4 2 + 10 2 + x 2 + y 2 + 12 2 + 14 2 + y 2 7 − 8 2 ⇒ x 2 + y 2 = 100 ∴ x = 8 , y = 6 : 1 , 2 , 3 , 4 , 5 , 6 = 1 − P ( ≥ 4 ) = 1 − 3 C 2 6 C 2 = 1 − 3 15 = 4 5 \begin{gathered} =8 \\ \frac{2+4+10+x+y+12+14}{7}=8 \\ \Rightarrow x+y=14 \\ =16 \\ 16=\frac{{2}^{2}+{4}^{2}+{10}^{2}+{x}^{2}+{y}^{2}+{12}^{2}+{14}^{2}+{y}^{2}}{7}-{8}^{2} \\ \Rightarrow {x}^{2}+{y}^{2}=100 \\ \therefore x=8,y=6 \\ :{1,2,3,4,5,6} \\ =1-P(\ge 4) \\ =1-\frac{{}^{3}{C}_{2}}{{}^{6}{C}_{2}} \\ =1-\frac{3}{15} \\ =\frac{4}{5} \end{gathered} = 8 7 2 + 4 + 10 + x + y + 12 + 14 = 8 ⇒ x + y = 14 = 16 16 = 7 2 2 + 4 2 + 10 2 + x 2 + y 2 + 12 2 + 14 2 + y 2 − 8 2 ⇒ x 2 + y 2 = 100 ∴ x = 8 , y = 6 : 1 , 2 , 3 , 4 , 5 , 6 = 1 − P ( ≥ 4 ) = 1 − 6 C 2 3 C 2 = 1 − 15 3 = 5 4 Answer the question to unlock