1in 6 chance
two dice are thrown once. write all the possible outcomes. find the probability of getting: (i) a total of 12 (ii) a total of 3 (iii) a total of 8
Prob = 4/36 = 1/9
It is 1/6. (It is just the same if the favorable outcome is 1,2,3,4 or 5, provided that it is a fair dice.)
To find the probability of getting an odd number at least once when a die is tossed thrice, we can use the complementary approach. The probability of not getting an odd number (i.e., getting an even number) in a single toss is ( \frac{3}{6} = \frac{1}{2} ). Therefore, the probability of getting an even number in all three tosses is ( \left(\frac{1}{2}\right)^3 = \frac{1}{8} ). Thus, the probability of getting an odd number at least once is ( 1 - \frac{1}{8} = \frac{7}{8} ).
5/6 = 83.3%
two dice are thrown once. write all the possible outcomes. find the probability of getting: (i) a total of 12 (ii) a total of 3 (iii) a total of 8
1/6
the probability is 2/6
Prob = 4/36 = 1/9
It is 1/6. (It is just the same if the favorable outcome is 1,2,3,4 or 5, provided that it is a fair dice.)
To find the probability of getting an odd number at least once when a die is tossed thrice, we can use the complementary approach. The probability of not getting an odd number (i.e., getting an even number) in a single toss is ( \frac{3}{6} = \frac{1}{2} ). Therefore, the probability of getting an even number in all three tosses is ( \left(\frac{1}{2}\right)^3 = \frac{1}{8} ). Thus, the probability of getting an odd number at least once is ( 1 - \frac{1}{8} = \frac{7}{8} ).
1/6 = 16.67%
5/6 = 83.3%
obviously it is a 50-50 chance
Zero
If it is a fair die that is rolled once, then the probability is 2/3.
The probability of getting a sum of 2 at least once is 0.8155