a box contains 20 cards bearing numbers 1,2,3,.........20. a card is drawn at random from the box. what is the propability that the number on the cards is (I) an odd number (ii) divisible by 2 or 3 (III) prime number (iv) not divisible by 10
Answers
Answer:
=1/2
Step-by-step explanation:
Total card=20
Prime number=(2,3,5,7,11,13,17,19)
=8
Probability of getting prime number=8/20
=2/5
Number which is neither divisible by 5 nor 10=(1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18,19)
=18
So probability=18/20
=9/10
Numbers which r even is=(2,4,6,8,10,12,14,16,18,20)
=10
So probability=10/20
=1/2
Answer:
Let E be event of drawing a card with number divisible by 2 or 3 from the cards with numbers 1 to 20
Numbers divisible by 2 or 3 from 1 to 20=2,3,4,6,8,9,10,12,14,15,16,18,20
No. of favorable outcomes=13
Total no. of possible outcomes =20
We know that, Probability P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
20
13
Therefore, the probability that a number on the card selected from numbers 1 to 20 divisible by 2 and 3 is
20
13
Solution(ii):
Let F be event of drawing a card with prime number from the cards with numbers 1 to 20
Prime numbers from 1 to 20=2,3,5,7,11,13,17,19
No. of favorable outcomes=8
Total no. of possible outcomes =20
We know that, Probability P(F) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
20
8
=
5
2
Therefore, the probability that a card with prime number is selected from numbers 1,2,3,...,20 =
5
2
Step-by-step explanation:
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