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Two number cubes are rolled.

What is the probability that the first lands on 6 and the second lands on an odd number?

Enter you answer, as a fraction in simplified form, in the box.

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MiNk2019314
The probability of two independent events both occurring is the product of their individual probabilities. The probability of rolling a 6 is 1/6 because there are 6 total numbers. The probability of rolling a odd number is 1/2, because there are an equal number of odd and even numbers on a die. Now multiply those probabilities together to get 1/6 * 1/2 = 1/12. 1/12 is the chance to roll a 6 and then roll an odd number.

The  probability that the first lands on 6 and the second lands on an odd number is [tex]1\div 12[/tex]

Calculation of the probabilities:

Since The probability of rolling a 6 should be 1/6 because there are 6 total numbers.

Now The probability of rolling an odd number is 1/2, since there are an equal number of odd and even numbers on a die.

Now the final probability should be

[tex]= 1\div 6 \times 1\div 2 \\\\= 1\div 12[/tex]

Therefore, The  probability that the first lands on 6 and the second lands on an odd number is [tex]1\div 12[/tex]

learn more about probability here: https://brainly.com/question/3066522