Math, asked by vanitashah019, 1 month ago

A box contains 100 boards which are numbered from 1 to 100. if one board at random from the box , find the probability that it bears,
1 ) an even prime numbers.
2) A number divisible by 7
3). A number with unit digit 9
give me this solutions​

Answers

Answered by shivrajswami0227
7

Answer:

There are 100 cards in the box marked with number 1 to 100.

A card is drawn randomly from it.

∴ Number of possible outcomes = 100

(1) Let A be the event that the card bears an even prime number.

There is only 1 even prime number from 1 to 100, that is '2'.

∴ The number of outcomes favourable to A is 1.

∴P(A)=

100

1

=0.01

(2) Let B be the event that the card bears a number divisible by 7.

Numbers divisible by 7 from 1 to 100 are

7,14,21,28,35,42,56,63,70,77,84,91,98

There are 13 multiples of 7 from the numbers 1 to 100.

∴ The number of outcomes favourable to B is 13.

∴P(A)=

100

13

=0.13

(3) Let be the event that the card bears a number that has 9 at the unit place.

The numbers from 1 to 100 that have 9 at the units place are

9,19,29,39,49,59,69,79,89,99

There are 10 numbers that have 9 at the unit place.

∴ The number of outcomes favourable to C is 10.

∴P(A)=

100

10

=0.1

Step-by-step explanation:

hope it help you have a nice day

Answered by HarshLikesAnime
2

Answer:

1) 1 in 100

2) 14 in 100 or 7 in 50

3) 1 in 100

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