A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number 1, 2, 3,…, 12 as shown in Fig. 13.3. What is the probability that it will point to:
(i) 10?
(ii) an odd number?
(iii) a number which is multiple of 3?
(iv) an even number?
Answers
1/12 , 1/2 , 1/3 , 1/2 are probabilities
Step-by-step explanation:
A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number 1, 2, 3,…, 12
Total Possible outcomes 12
S = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9, 10 , 11 , 12}
n(S) = 12
(i) 10
n(A) = 1
Number of Favorable outcome = 1
Probability = 1/12
(ii) an odd number
B = {1 , 3 , 5 , 7 , 9 , 11 }
n(B) = 6
Probability = 6/12 = 1/2
(iii) a number which is multiple of 3
C = { 3 , 6 , 9 , 12 }
n(C) = 4
Probability = 4/12 = 1/3
(iv) an even number
D = { 2 , 4 , 6 , 8 , 10 , 12}
n(D) = 6
Probability = 6/12 = 1/2
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Answer:
Ans. Out of 8 numbers, an arrow can point any of the numbers in 8 ways.
Total number of favourable outcomes = 8
(i) Favourable number of outcomes = 1
Hence, P (arrow points at 8) =
(ii) Favourable number of outcomes = 4
Hence, P (arrow points at an odd number) =
(iii) Favourable number of outcomes = 6
Hence, P (arrow points at a number > 2) =
(iv) Favourable number of outcomes = 8
Hence, P (arrow points at a number < 9) = =1