Question

the faces of a die bear numbers 0,1,2,3,4,5. if the die is rolled twice, then find the probability that the product of digits on the upper face is zero....... pls ans in full method.. I'll mark them as BRAINLIEST​

Answer 1

Step-by-step explanation:

No. of different number written on the dice = 6

No. of zeros written on the dice = 1 Then, probability of getting a zero once = 1/6

So, probability of getting zero when dice is twice rolled = 1/6 * 2/2 = 2/12

So the answer is 2/12 which can be simplified to get 1/6.

Answer 2

Answer

Here we have>

No. of possible outcomes(s) =>

S= (0,0) (0,1),(0,2),(0,3),(0,4),(0,5),(1,0),(1,1),(1,2),(1,3),(1,4),(1,5),(2,0),(2,1),(2,2),(2,3),(2,4),(2,5),(3,0),(3,1),(3,2),(3,3),(3,4),(3,5),(4,0),(4,1),(4,2),(4,3),(4,4),(4,5),(5,0),(5,1),(5,2),(5,3),(5,4),(5,5).

n(total outcomes) = 36

Now, let the number of favorable outcomes i.e the product of two numbers are zero be =>

n(getting product as zero)

=(0,0) (0,1),(0,2),(0,3),(0,4),(0,5),(1,0) ,(2,0),(3,0),(4,0),(5,0).

i.e n(getting product as zero) = 11.

Hence the probability that the product of digits on the upper face is zero will be

P(getting the product as zero)

=$\dfrac{n(getting product as zero)}{n(total outcomes)}$

=$\dfrac{11}{36}$

Hence the probability of getting the product of digits on the upper face is zero is $\dfrac{11}{36}$.

Remember

1)Probability = $\dfrac{number of favourable outcomes}{total number of outcomes}$.

2)P(favorable outcomes) + P(unfavourable outcomes) = 1