Question

A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].

Answer 1

Hello!

Total number of events = $6 \times 6 = \bf{36}$

$\underline{\sf EVENTS \:OF\:GETTING \:5}$ :

• (5, 1)
• (5, 2)
• (5, 3)
• (5, 4)
• (5, 5)
• (5, 6)
• (1, 5)
• (2, 5)
• (3, 5)
• (4, 5)
• (6, 5)

Then,

P (getting number 5) = $\red{\bf{\dfrac{11}{36}}}$

P (not getting number 5) = $1 - \dfrac{11}{36} = \red{\bf{\dfrac{25}{36}}}$

P (getting number 5 once) = $\red{\bf{\dfrac{10}{36}}}$

Cheers!

Answer 2

Answer:

Step-by-step

Total event =6×6 =36

Events of getting 5 are = (5,1),( 5,2),( 5,3),( 5,4),( 5,5),( 5,6),(1,5),(2,5),(3,5), (4,5),( 6,5)

probability of getting getting 5 =11/36

Probability of not getting 5 =1-(11/36) =25/36

probability of getting. 5 exactly one time is =10/36