Given: Two dice are thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur
$(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),$
$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),$
$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),$
$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),$
$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),$
$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$
Favorable events i.e. 5 will not come up on either of them
$(1, 1), (1, 2), (1, 3), (1, 4), (1, 6),$
$(2, 1), (2, 2), (2, 3), (2, 4), (2, 6),$
$(3, 1), (3, 2), (3, 3), (3, 4), (3, 6),$
$(4, 1), (4, 2), (4, 3), (4, 4), (4, 6),$
$(6, 1), (6, 2), (6, 3), (6, 4), (6, 6)$
Hence total number of favorable events i.e. $5$ will not come up on either of them is $25$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability that 5 will not come up on either of them is equal to $=\frac{25}{36}$
- Favorable events i.e. 5 will come on at least once
$(1, 5), (2, 5), (3, 5), (4, 5),$
$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5),$
Hence total number of favorable events i.e.$5$ will not come on at least once is $11$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting i.e. $5$ will come on at least once is equal to $=\frac{11}{36}$
- Favorable events i.e. $5$ will come on both side is $(5, 5)$
Hence total number of favorable events i.e. $5$ will come on both side is $1$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting $5$ will come on both side is equal to $=\frac{1}{36}$
