Question

A suitcase can be locked using a 3 digit number.Whats the probability of opening lock.If its known that one of the number of the combination is 3

Answer 1

Heya friend,

Here we go :

Answer:

$\frac{1}{100}$

Step-by-step explanation:

No. of ways in which we can open the box =

__ ×  __  ×  __

Here,

For the first number, we have 10 ways varying from (0-9) and as it is not mentioned that whether the repetition is allowed or not, we'll take the repetition in consideration.

So, for the second number too, we have 10 ways

and,

for the third number we have only one way as we know one of the number of the combination is 3.

So.

Total ways =

⇒ 1 × 10 × 10

⇒ 100 ways

Now,

the favourable case would be 1 only as there would be only 1 combination in which the lock would open.

So,

Probability = $\frac{favourable}{total}$

                  = $\frac{1}{100}$

Thanks !

#BAL #answerwithquality

Answer 2

GIVEN:-

A suitcase can be locked using a 3 digit number.

TO FIND:-

Whats the probability of opening lock. If its known that one of the number of the combination is 3.

SOLUTION:-

As per given data,

the lock is locked by three digits.

And condition is that minimum one digit is 3.

So,

taking three box for three digits.

i)      ___, ___, ____.

         1       9       9

above we fixed 3 at a single box and 9 digits can be fill to two box

therefore, total combination of number = 1 × 9 × 9

                                                                  = 81.

ii)    ____, ____, ____.

       9           1         9

above we fixed 3 at a second box and 9 digits can be fill to two box

therefore, total combination of number = 9× 1 × 9

                                                                  = 81.

iii)  ____, ____, ____.

       9           9       1

above we fixed 3 at a third box and 9 digits can be fill to two box

therefore, total combination of number = 9× 9 × 1

                                                                  = 81.

   From the above combination total outcomes = 81 + 81 + 81

                                                                               = 243.

Now ,

Probability = $\frac{possible ~outcomes}{total ~outcomes}$    ( possible outcomes = 1 )

                  = $\frac{1}{243}$  

Therefore, probability = $\frac{1}{243}$.

THANKS.

#SPJ2.

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