Total number of slips = 25 + 50 = 75
Number of slips marked with Rs 1 = 19 + 45 = 64
\(\therefore\) Number of slips marked other than 1 = 75 - 64 = 11
\(\therefore\) Required probability  = \(\frac{11}{75}\)

Possible outcomes 
(0,0),(0,1),(0,1),(0,1),(0,6),(0,6)
(1,0),(1,1),(1,1),(1,1),(1,6),(1,6)
(1,0),(1,1),(1,1),(1,1),(1,6),(1,6)
(1,0),(1,1),(1,1),(1,1),(1,6),(1,6)​​​​​​​
(6,0),(6,1),(6,1),(6,1),(6,6),(6,6)
(6,0),(6,1),(6,1),(6,1),(6,6),(6,6)​​​​​​​
Different total scores are 0, 1, or 12
Let A = getting a total of 7
No. of favourable outcomes are = 12
\(\therefore\) P(A) = \(\frac{12}{36}=\frac{1}{3}\)​​​​​​​

Total number of phones = 48
Let A = phone is good
Number of good phones = 42
\(\therefore\) P(A) =\(\frac {42}{48}=\frac {7}{8}\)
\(\therefore\) Probability that Varnika will buy a phone =\(​​\frac {7}{8}\) 
Let B = Phone has no major defect number of Phones having no major defects = 48 - 3 = 45 
\(\therefore\) P(B)=\(\frac {45}{48}=\frac {15}{16}\)
Probability that phone is acceptable to the trader = \(\frac {15}{16}\)

Total number of bulbs=24
Number of defective buls=6
Number of good ones=24-6=18
P (not defective)=\({18\over24}={3\over4}\)
P (2nd bulb is defective) =\(5\over23\)
[first bulb is defective and not replaced]

Number of total outcomes, n(S)=36
(i) When product of the numbers on the top of the dice is 6.
So, the possible ways are {(1,6), (2,3), (3,2), (6,1).}
Number of possible ways, n(E₁)=4
∴ Required probability\(=\frac { n(E_{ 1 }) }{ n(S) } =\frac { 4 }{ 36 } =\frac { 1 }{ 9 } \)
(ii) When the product of the numbers on the top of the dice is 12.
So, the possible ways are {(2,6), (3,4), (4,3), (6,2)}
Number of possible ways, n(E₂)=4
∴ Required probability\(=\frac { n(E_{ 2 }) }{ n(S) }\)
\( =\frac { 4 }{ 36 } =\frac { 1 }{ 9 } \)
(iii) Product of the numbers on the top of the dice cannot be7. its probability is zero