What is the probability that a prime number selected at random from numbers 1 2 3 35?
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Total no. of possible outcomes = 35 {1, 2, 3, …. 35} Show
E ⟶ event of getting a prime no. No. of favourable outcomes = 11 {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31} Probability, P(E) =`"No.of favorable outcomes"/"Total no.of possible outcomes"` = 11/35 Solution: Events in probability are characterized as certain probable outcomes of an experiment that make up a subset of a finite sample space. Every event will always have a probability of occurrence between 0 and 1. Numbers from 1, 2, 3….. 35 are a total of 35. Total no. of possible outcomes = 35 Let E = event of getting a number which is a multiple of 7 No. of favourable outcomes = 5 {7, 14, 21, 28, 35} Probability, P(E) = Number of favourable outcomes/ Total number of outcomes P(E) = 5/35 = 1/7 Given: Numbers \( 1,2,3, \ldots, 35 \) are given. To do: We have to find the probability that a number selected at random from the numbers \( 1,2,3, \ldots, 35 \) is a prime number. Solution: Numbers \( 1,2,3, \ldots, 35 \) are given. This implies, The total number of possible outcomes $n=35$. Prime numbers from 1 to 35 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31. Total number of prime numbers from 1 to 35 $=11$ Total number of favourable outcomes $=11$. We know that, Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$ Therefore, Probability that a number selected from the numbers \( 1,2,3, \ldots, 35 \) is a prime number $=\frac{11}{35}$ The probability that a number selected from the numbers $1, 2, 3, ........, 35$ is a prime number is $\frac{11}{35}$.
Updated on 10-Oct-2022 10:55:12
Total no. of possible outcomes = 35 {1, 2, 3, …. 35} (i) E ⟶ event of getting a prime no. No. of favourable outcomes = 11 {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31} Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 11/35 (ii) E ⟶ event of getting no. which is multiple of 7 No. of favourable outcomes = 5 {7, 14, 21, 28, 35} P(E) = 5/35 = 1/7 (iii) E ⟶ event of getting no which is multiple of 3 or 5 No. of favourable outcomes = 16 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 5, 10, 20, 25, 35} P(E) = 16/35 Given: Numbers \( 1,2,3, \ldots, 35 \) are given. To do: We have to find the probability that a number selected at random from the numbers \( 1,2,3, \ldots, 35 \) is a prime number. Solution: Numbers \( 1,2,3, \ldots, 35 \) are given. This implies, The total number of possible outcomes $n=35$. Prime numbers from 1 to 35 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31. Total number of prime numbers from 1 to 35 $=11$ Total number of favourable outcomes $=11$. We know that, Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$ Therefore, Probability that a number selected from the numbers \( 1,2,3, \ldots, 35 \) is a prime number $=\frac{11}{35}$ The probability that a number selected from the numbers $1, 2, 3, ........, 35$ is a prime number is $\frac{11}{35}$. What is the probability that a number selected from numbers 1 2 3?P=3010=31.
What is the probability that a number selected from numbers 1 2 3 30 is prime number?Solution : Prime numbers from 1 to 30 are 2,3,5,7,11,13,17,19,23,29.
Their number is 10. ` :. ` P(getting a prime number) =` 10/30 = 1/3`. What is the probability that a random number is prime?The prime number theorem implies that the probability that a random number n is prime, is about 1/log n (technically, the probability a number m chosen from the set {1,2,...,n} is prime is asymptotic to 1/log n).
What is the probability that a number selected from 1 2 3 25 is a prime number when each of the given number is equally likely to be selected?Hence, the Probability of getting a prime number from 1 to 25,P(E) is 9/25. HOPE THIS ANSWER WILL HELP YOU….
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