Question

3. A die is thrown once. What is the probability of getting: i) a prime number? ii) a number greater than 2? Iii) a number other than 2 and 5?

Answer 1

Required Answer:

i) a prime number

⇒ $\bigstar \: \boxed{\sf {P(E) = \dfrac{n(E)}{n(S)} }}$

>> P(E) = Probability of the occurrence of the event.

>> n(E) = Number of favourable outcomes

>> n(S) = Number of possible outcomes

• Total number of possible outcomes = {1,2,3,4,5,6} = 6

• Number of favourable outcomes = {2,3,5} = 3

(Since, 2,3,5 are prime numbers upto 6)

⇒$\sf { P(E) = \dfrac{n(E)}{n(S)} }$

⇒$\sf { P(a \: prime \: number) = \dfrac{3}{6} }$

⇒$\sf\red { P(a \: prime \: number) = \dfrac{1}{2} }$

⠀⠀⠀⠀⠀_____________

ii) a number greater than 2

⇒ $\bigstar \: \boxed{\sf {P(E) = \dfrac{n(E)}{n(S)} }}$

>> P(E) = Probability of the occurrence of the event.

>> n(E) = Number of favourable outcomes

>> n(S) = Number of possible outcomes

• Total number of possible outcomes = {1,2,3,4,5,6} = 6

• Number of favourable outcomes = {3,4,5,6} = 4

⇒$\sf { P(E) = \dfrac{n(E)}{n(S)} }$

⇒$\sf { Probability = \dfrac{4}{6} }$

⇒$\sf\red { Probability= \dfrac{2}{3} }$

⠀⠀⠀⠀⠀_____________

ii) a number other than 2 and 5

⇒ $\bigstar \: \boxed{\sf {P(E) = \dfrac{n(E)}{n(S)} }}$

>> P(E) = Probability of the occurrence of the event.

>> n(E) = Number of favourable outcomes

>> n(S) = Number of possible outcomes

• Total number of possible outcomes = {1,2,3,4,5,6} = 6

• Number of favourable outcomes = {1,3,4,6} = 4

⇒$\sf { P(E) = \dfrac{n(E)}{n(S)} }$

⇒$\sf { Probability = \dfrac{4}{6} }$

⇒$\sf\red { Probability= \dfrac{2}{3} }$

⠀⠀⠀⠀⠀_____________

Answer 2

P(E) = (No. of E outcomes)/(Total outcomes)

i) Number of prime numbers it can encounter with = 3 (2, 3, 5)

Total number of possible outcomes = 6

∴ P(prime no.) = (No. of prime numbers)/(Total outcomes)

⇒ P(prime no.) = 3/6 = 0.5

ii) Numbers greater than 2 = 4

No. of possible outcomes = 6

∴ P(> 2) = (Numbers greater than 2 in a dice)/(Total number of possible outcomes)

⇒ P(> 2) = 4/6 = ⅔

iii) Numbers other than 2 and 5 = 4

No. of possible outcomes = 6

⇒ P(other than 2 and 5) = 4/6 = ⅔.