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In a single throw of two dice, what is the probability of getting odd number on one die and 6 on the other?
Please explain!
1/12
Dear student,
Favourable cases are 3
Total=36
Required probabilty=3/36=1/12
S= {(1,1)(1.2)(1,3)(1,4)(1,5)(1,6).......(6,6)} = 36.
Possible outcomes are 1,3,5 on 1st dice and 6 on other.
= {(1,6)(3,6)(5,6)} = 3.
Probability= 3 /36. = 1/12
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