Three dice are thrown simultaneously. The probability of getting a total of atleast 5 of the numbers appearing on their tops is?

Asked by adityaahuja099 | 12th Dec, 2017, 05:33: PM

Expert Answer:

begin mathsize 16px style Three space dice space are space thrown space simultaneously comma space total space number space of space outcomes equals 6 cubed equals 216
Getting space straight a space total space of space at space least space 5 space means space 5 space and space more space than space 5.
Now comma space Outcomes space with space total space less space than space 5 equals left parenthesis 1 comma space 1 comma space 1 right parenthesis comma space left parenthesis 1 comma space 1 comma space 2 right parenthesis comma space left parenthesis 1 comma space 2 comma space 1 right parenthesis comma space left parenthesis 2 comma space 1 comma space 1 right parenthesis
Number space of space outcomes space with space total space less space than space 5 equals 4
Thus comma space probability space of space getting space straight a space total space less space than space 5 equals 4 over 216
Hence comma space required space probability space of space getting space straight a space total space of space at space least space 5
equals 1 minus straight P left parenthesis getting space straight a space total space less space than space 5 right parenthesis
equals 1 minus 4 over 216
equals fraction numerator 216 minus 4 over denominator 216 end fraction
equals 212 over 216
equals 53 over 54 end style

Answered by Rashmi Khot | 13th Dec, 2017, 08:46: AM