TG EAMCET 2 May 2025 Shift 2 Paper

Given, matrix P=[

1	2	3
4	5	6
7	8	9

]
Odd elements ={1,3,5,7,9}
Total =5
Even elements ={2,4,6,8}
Total =4
Case I 3 odd numbers numbers of ways =‌5C3=10
Case II 1 odd and 2 even numbers of ways =‌5C1×‌4C2
=5×6=30
Total favourable cases for A
=10+30=40
And total ways for selecting 3 elements from 9 elements =‌9C3=84
∴‌‌P(A)=‌

40

84

Selecting 3 elements in a row or column.
3 rows ×1 way =3 ways
3 column ×1 way =3 ways
total =6 ways
Now, Row 1:1+2+3=6 (even)
Row 2 : 4+5+6=15 (odd)
Row 3:7+8+9=24 (even)
Column I : 1+4+7=12 (even)
Column II : 2+5+8=15 (odd)
Column III : 3+6+9=18 (even)
‌∴P(A∩B)=‌

2

84

‌‌‌P(B)=‌

6

84

‌∴‌‌P(‌

A

B

)=‌

P(A∩B)

P(B)

=‌

‌ 2 84

6 84

=‌

1

3

‌∴P(A)+P(‌

A

B

)‌=‌

40

84

+‌

1

3

‌=‌

40+28

84

‌=‌

68

84

=‌

17

21