20. Consider olling a fair die. Answer the following questions. Here let S = set

Question

20. Consider olling a fair die. Answer the following questions. Here let S = set of all posisble outcomes when we roll a die. Therefore S (1,2,3,4,5,6)n(S) -6 a. Odds in favor of getting a number less than 2 (Ans: 1:5 n(E)-1 Here let E = set of nambers less than 2. Therefore E-(1) and therefore,n(E) n(S)-n(E) 6-1 5 Odds(E)-n(E): n(E') = 1:5 Odds in tavor of getting a number less than 4 [Ans: 1:11 Here let E b. set of numbers less than 4. Therefore E = (1,23)-n(E-3 and therefore, n(E)-n(S)-n(E) = 6-3 = 3. Odds(E) = n(E); n(E')s 3:3 = 1 1 c. Odds against getting a number more than 5 [Ans: 5:1] d. Odds of getting a prime number lAns: 1:1] e. Odds against getting a multiple of three. [Ans: 2:1] t. Odds of getting an even prime [Ans: 1:5 g. Probability of getting an even prime lAns: 0.1667) h. Probability of getting a multiple of 4 [Ans: 0.1667) i. Probability of getting a number divisible by 1.75 [Ans: 01 Venkateswara Rao Mudhonara all so B Exam a Review Sheet

Explanation / Answer

(c) E=more than 5={6}, n(E)=1,

so n(E')=n(S)-n(E)=6-1=5

odds against E=n(E')/n(E)=5/1=5:1

(d) E=prime number={2,3,5}, n(E)=3,

so n(E')=n(S)-n(E)=6-3=3

odds of geeting E=n(E)/n(E')=3/3=1:1

(e) E=multiple of 3={3,6}, n(E)=2, n(E')=4

odds against=n(E')/n(E)=4/2=2/1=2:1

(f)E=even prime number={2},n(E)=1,n(E')=5

odds against=n(E')/n(E)=5/1=5:1

(g) E=even prime number={2},n(E)=1

P(E)=n(E)/n(S)=1/6=0.1667

(h)E=multiple of 4={4},n(E)=1

P(E)=n(E)/n(S)=1/6=0.1667

(i)E=number divisible by 1.75

E={ }, n(E)=0

P(E)=n(E)/n(S)=0/6=0

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