MHT CET · Maths · Probability
One ticket is selected at random from 50 tickets numbered \(\{00,01,02, \ldots . ., 49\}\). Then the probability that the sum of the igits on the selected ticket is 8 , given that the product of these digits is zero, is
- A \(\frac{1}{50}\)
- B \(\frac{1}{14}\)
- C \(\frac{14}{50}\)
- D \(\frac{1}{10}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{14}\)
Step-by-step Solution
Detailed explanation
\(S=\{00,01,02, \ldots \ldots ., 49\}, n(S)=50 \)
\( E_1=\{08,17,26,35,44\}, n\left(E_1\right)=5 \)
\( E_2=\{00,01,02,03,04,05,06,07,08,09,10,20,30,\) \(40\},n\left(E_2\right)=14 \)
\( E_1 \cap E_2=\{08\}, n\left(E_1 \cap E_2\right)=1 \)
\( \text { Required probability }=p\left(E_1 / E_2\right)=\frac{n\left(E_1 \cap E_2\right)}{n\left(E_2\right)}=\frac{1}{14}\)
\( E_1=\{08,17,26,35,44\}, n\left(E_1\right)=5 \)
\( E_2=\{00,01,02,03,04,05,06,07,08,09,10,20,30,\) \(40\},n\left(E_2\right)=14 \)
\( E_1 \cap E_2=\{08\}, n\left(E_1 \cap E_2\right)=1 \)
\( \text { Required probability }=p\left(E_1 / E_2\right)=\frac{n\left(E_1 \cap E_2\right)}{n\left(E_2\right)}=\frac{1}{14}\)
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