- \(\frac{1}{6}\)
- \(\frac{1}{3}\)
- \(\frac{5}{6}\)
- \(\frac{1}{9}\)

Option 1 : \(\frac{1}{6}\)

With hundreds of Questions based on Probability, we help you gain expertise on Quantitative Aptitude. All for free. Explore Testbook Learn to attain the subject expertise with us.

**GIVEN:**

One dice shows a multiple of 3.

Other dice shows even number.

**CONCEPT:**

Total number of outcomes in two dice is 36.

**FORMULA USED:**

P = Favorable outcomes/Total outcomes

**CALCULATION:**

There are only 6 such cases as required,

(3,2), (3,4) (3,6) (6,2) (6,4) (6,6)

∴ Required probability = 6/36 = 1/6

**∴ The probability is 1/6.**

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