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Given
    • Two dice, each numbered 1 to 6, are tossed.

To find

    • The probability that neither shows a "4"

Approach
    • Probability that neither shows a "4" = P(First dice does not show 4) × P(Second dice does not show 4)
      o P(Dice does not show 4) = \(\frac{Total\ cases\ of\ a\ dice\ not\ showing\ 4 }{ Total\ cases}\)
      o P(Dice does not show 4) = \(\frac{5 }{ 6}\)
    • Probability that neither shows a "4" = \(\frac{5 }{ 6}\) × \(\frac{5 }{ 6}\) = \(\frac{25 }{ 36}\)

Hence, the correct answer is option D.
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Two dice, each numbered 1 to 6, are tossed.

As we are rolling 2 dice => Number of cases = \(6^2\) = 36

What is the probability that neither shows a "4" ?

P(Neither shows 4) = P(First die does not show a 4) * P(Second die does not show a 4) = \(\frac{5}{6}\) * \(\frac{5}{6}\) = \(\frac{25}{36}\) [ As we can get any number out of 4 in both the dice ]

So, Answer will be D
Hope it helps!

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