TS EAMCET · Maths · Probability
Keep fixed first position whose possible numbers are 1, \[ \begin{aligned} & 2,3 \text { and } 4 . \ & =1 \times 4+1 \times 4+1 \times 4+1 \times 3=15 . \ & \mathrm{P}(\mathrm{E}) \frac{\text { Favourable }}{\text { Total }}=\frac{15}{36}=\frac{5}{12} \end{aligned} \] So, option (d) is correct.
- A \(\frac{94}{663}\)
- B \(\frac{62}{663}\)
- C \(\frac{30}{663}\)
- D \(\frac{64}{663}\)
Answer & Solution
Correct Answer
(B) \(\frac{62}{663}\)
Step-by-step Solution
Detailed explanation
Given a well shuffled pack of 52 cards.…
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