Question

A box contains slips having numbers from 30 to 99 written on it
pencil box for a slip having multiple of 3, a pen for digit greater than 70 on slip
and a lunch box for having a slip with multiple of 1
out of the box
wers from 30 to 99 written on it. there is prize of
**tiple of 3, a pen for a digt greater than 70 on slip
aving a slip with multiple of 10 written on it. Jaya picks a slip
out of the box.​

Answer 1

(1) The probability of winning a pencil box = $\frac{24}{70}$ = $\frac{12}{35}$  = 34.28%

(2) The probability of winning a pen = $\frac{29}{70}$ =  41.42%

(3) The probability of winning a pencil box = $\frac{7}{70}=\frac{1}{10}$  = 10%

Step-by-step explanation:

A box contains slips having numbers from 30 to 99 written on it.

Total number of slips in the box = (99 - 29) = 70 slips

There is a prize of a pencil box for a slip having multiple of 3.

Number of slips having multiple of 3. This can be found with arithmetic progression :

$a_{n}=a+(n-1)d$  

$99=30(n-1)3$

$69=(n-1)3$

23 = n - 1

n = 24

There are 24 slips having digits that are multiple of 3.

So, the probability of winning a pencil box = $\frac{24}{70}$ = $\frac{12}{35}$  = 34.28%

There is a prize of a pen for a slip having digit greater than 70.

There are slips having digits greater than 70 = 99 - 70 = 29

So the probability of winning a pen = $\frac{29}{70}$ =  41.42%

There is a prize of a lunch box for a slip having digit with multiple of 10.

there are number of slips having digit of multiple of 10 = 30,40,50,60,70,80,90 = 7 slips

So the probability of winning a pencil box = $\frac{7}{70}=\frac{1}{10}$  = 10%

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