A combination lock on a suitcase has 3 wheels, each labelled with nine digits from 1 to 9. If an opening combination is a particular sequence of three digits with no repeats, what is the probability of a person guessing the right combination ?

Open in App

Solution

Total Number of ways in which we can select a combination (without repeatition) = $^{9} C_{1} \times^{8} C_{1} \times^{7} C_{1} = 9 \times 8 \times 7 = 504$

Out of all these combinations, only 1 is correct.

Hence, p(E) = $\frac{1}{504}$

Suggest Corrections

Similar questions

Q. The number lock of a suitcase has

4

wheels each labelled with ten digits i.e. from

0

9

. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

Q. A number lock on a suitcase has

3

wheels each labelled with

10

digits from

0

9

. If opening of the lock is a particular sequence of three digits with no repeats, how many such sequences will be possible?

Q. A number lock on a suitcase has

3

wheels each labeled with

10

digits from

0

9

. If the opening of the lock is a particular sequence of three digits with no repeats, how many such sequences will be possible?

Q. A number of lock on a suitcase has three wheels each labelled with ten digits

0

9

. The opening of the lock is a particular sequence of three digits with no repeats. If a person tries to open the lock, then the number of unsuccessful attempts made by him is

A number lock on a suitcase has 3 wheels each labelled with ten digits 0 to 9. If opening of the lock is a particular sequence of three digits with no repeats, how many such sequences will be possible? Also find the number of unsuccessful attempts to open the lock.

Join BYJU'S Learning Program

A combination lock on a suitcase has 3 wheels, each labelled with nine digits from 1 to 9. If an opening combination is a particular sequence of three digits with no repeats, what is the probability of a person guessing the right combination ?

Total Number of ways in which we can select a combination (without repeatition) = 9C1 × 8C1 × 7C1= 9×8×7=504 Out of all these combinations, only 1 is correct. Hence, p(E) = 1504

Total Number of ways in which we can select a combination (without repeatition) = $^{9} C_{1} \times^{8} C_{1} \times^{7} C_{1} = 9 \times 8 \times 7 = 504$

Out of all these combinations, only 1 is correct.

Hence, p(E) = $\frac{1}{504}$