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Mathematics

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5

(iv) a prime number less than 30.

Probability

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Answer

A disc is drawn at random from the box means that all the outcomes are equally likely.

Sample space = {1, 2, 3, ….., 90}, which has 90 equally likely outcomes.

(i) Let E1 be the event of drawing a disc with two digit number.

E1 = {10, 11, 12, 13, ……, 90}.

∴ The number of favourable outcomes to the event E1 = 81.

P(E1)=No. of favourable outcomes to E1Total no. of possible outcomes=8190=910.\therefore P(E1) = \dfrac{\text{No. of favourable outcomes to } E1}{\text{Total no. of possible outcomes}} = \dfrac{81}{90} = \dfrac{9}{10}.

Hence, the probability of drawing a disc with two digit number is 910.\dfrac{9}{10}.

(ii) Let E2 be the event of drawing a disc with perfect square number.

E2 = {1, 4, 9, 16, 25, 36, 49, 64, 81}.

∴ The number of favourable outcomes to the event E2 = 9.

P(E2)=No. of favourable outcomes to E2Total no. of possible outcomes=990=110.\therefore P(E2) = \dfrac{\text{No. of favourable outcomes to } E2}{\text{Total no. of possible outcomes}} = \dfrac{9}{90} = \dfrac{1}{10}.

Hence, the probability of drawing a disc with perfect square number is 110\dfrac{1}{10}.

(iii) Let E3 be the event of drawing a disc with number divisible by 5.

E3 = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}.

∴ The number of favourable outcomes to the event E3 = 18.

P(E3)=No. of favourable outcomes to E3Total no. of possible outcomes=1890=15.\therefore P(E3) = \dfrac{\text{No. of favourable outcomes to } E3}{\text{Total no. of possible outcomes}} = \dfrac{18}{90} = \dfrac{1}{5}.

Hence, the probability of drawing a disc with number that is divisible by 5 is 15\dfrac{1}{5}.

(iv) Let E4 be the event of drawing a disc with prime number less than 30.

E4 = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}.

∴ The number of favourable outcomes to the event E4 = 10.

P(E4)=No. of favourable outcomes to E4Total no. of possible outcomes=1090=19.\therefore P(E4) = \dfrac{\text{No. of favourable outcomes to } E4}{\text{Total no. of possible outcomes}} = \dfrac{10}{90} = \dfrac{1}{9}.

Hence, the probability of drawing a disc with prime number less than 30 is 19\dfrac{1}{9}.

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