Question

2. A die is thrown. Find the probability of getting
(1) a prime number INCERTI
fii) 2 or 4
(iii) a multiple of 2 or 3
(iv) an even prime number (CBSE 2008]
(v)a number greater than 51CBSE 2008] (vi) a number lying between 2 and 6 photo answer

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Answer 1

Answer:

Given : A dice is thrown once.

A die has 6 faces marked as 1, 2, 3, 4, 5 and 6.

Possible outcomes are 1, 2, 3, 4, 5 and 6 = (6).

Total number Possible outcomes  6.

(i) Let E1 = Event of getting a prime number  

Prime number on a dice are 2,3,5  

Number of outcomes favourable to E1 =  3

Probability (E1) = Number of favourable outcomes / Total number of outcomes

P(E1) = 3/6 = ½  

P(E1) = 1/2

Hence, probability of getting a prime number ,P(E1)  = ½.

(ii) Let E2 = Event of getting 2 & 4  

Number of outcomes favourable to E2 =  2

Probability (E2) = Number of favourable outcomes / Total number of outcomes

P(E2) = 2/6 = 1/3

P(E2) = 1/3

Hence, probability of getting 2 & 4,  P(E2)  = ⅓..

(iii) Let E3 = Event of getting a multiple of 2 or 3  

Multiple of 2 or 3  on a dice are 2,3,4,6

Number of outcomes favourable to E3 =  4

Probability (E3) = Number of favourable outcomes / Total number of outcomes

P(E3) = 4/6 = 2/3

P(E3) = 2/3

Hence, probability of getting a multiple of 2 or 3 , P(E3)  = ⅔ .

(iv) Let E4 = Event of getting an even prime number  

Even Prime number on a dice are 2

Number of outcomes favourable to E4 =  1

Probability (E4) = Number of favourable outcomes / Total number of outcomes

P(E4) = 1/6  

Hence, probability of getting an even prime number, P(E4)  = ⅙ .

(v) Let E5 = Event of getting a number greater than 5

A number greater than 5 on a dice is 6

Number of outcomes favourable to E5 = 1

Probability (E5) = Number of favourable outcomes / Total number of outcomes

P(E5) = 1/6  

Hence, probability of getting a number greater than 5 , P(E5)  = ⅙ .

(vi) Let E6 = Event of getting a number lying between 2 and 6  

Number lying between 2 and 6 on a dice are 3,4,5  

Number of outcomes favourable to E6 =  3

Probability (E6) = Number of favourable outcomes / Total number of outcomes

P(E6) = 3/6 = ½  

P(E6) = 1/2

Hence, probability of getting a number lying between 2 and 6, P(E6)  = ½.