1hey help yr
1. two integers x and y ar drawn (without replacement) out of set (0,1,2,3,---,10) find the probability that -5<x-y<5?
2. Two natural numbers ( x and y) ar choosen at random. What is the probability that x^2 + y^2 is divisible by
a) 10
b) 7
3. Each coefficient in equation lx^2 + mx + n = 0, is determined by throwing a symmetrical cube, What is the probability that the equation will have real roots?
4. If four squares are choosen at random from chess board? Find the probability that they lie on diagnol line?
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UP 0 DOWN 0 0 10
10 Answers
14) i think simply
15C464C4
3) for real roots
l2≥4mn
l,m,n can each range from 1-6 ( i assume that;s what you mean by cube)
find all the cases manually
1ther are two diagonal lines giving 15 squares which can be on the any of the two diagonal lines.
23widout replacement implies x≠y
-5 + y ≤ x ≤ 5 + y
y = 0 ,
5 ≥ x ≥ -5 ( 5 values of x )
y =1 ,
6 ≥x ≥ -4 ( 6 values )
y =2 ,
7 ≥x ≥ -3 ( 7 values )
y =3 ,
8 ≥x ≥ -2 ( 8 values )
y =4 ,
9 ≥x ≥ -1 ( 9 values )
y =5 ,
10 ≥x ≥ 0 ( 10 values )
y =6 ,
11 ≥x ≥ 1 ( 9 values )
similarly
y = 7 , ( 8 values )
y = 8 ( 7 values )
y = 9 ( 6 values )
y = 10 ( 5 values )
so total pairs satisfying the inequality = 2(5 +6+7+8+9) + 10 = 80
total possible = 110
so p = 8/11 ?
12) let x = 10m + λ
y = 10n + μ, where λ,μ ε[0,9]
x2+y2 = 10k +( λ2 + μ2)
clearly x2+y2 is divisible by 10 if λ2 + μ2 is divisible by 10
so possible cases are (1,3)(3,1)(2,4)(4,2)(2,6)(6,2)(4,8)(8,4)(5,5)(6,8)(8,6)(9,3)(3,9)
so reqd probability = 13/81
similarly do for 7
(check the ans if i might have missed out any other possibilities)
14) check out the diagonal lines...
4 squares can be chosen from the diagonal with atleast 4 sqres..
so starting from one corner towards other we can select 4 sqres from each diagonal in
4C4 +5 C4 + 6C4 + 7C4 + 8C4 + 7C4 + 6C4 + 5 C4 + 4C4
similar case observed when approached from another corner to its opposite....
so reqd probability = 4(4C4 +5 C4 + 6C4 + 7C4) + 28C464C4