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(B) There are $10$ possible digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.Since the last two digits are different,the total number of ways to choose and arra

Vedclass · Accepted Answer

$\frac{1}{90}$

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(B) There are $10$ possible digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.Since the last two digits are different,the total number of ways to choose and arrange these two digits is given by the permutation formula ${}^{10}P_2$.${}^{10}P_2 = 10 	imes 9 = 90$.Out of these $90$ possible outcomes,only $1$ outcome is the correct sequence of the last two digits.Therefore,the required probability is $\frac{1}{90}$.

While dialing a telephone number,an old man forgets the last two digits,remembering only that they are different. If he dials the last two digits at random,what is the probability that the number is dialed correctly?

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