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One ticket is selected at a random from 50 ticket numbered 00, 01, 02, 03, ………………, 49.Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals,
A.$\dfrac{1}{14}$
B.$\dfrac{1}{7}$
C.\[\dfrac{5}{14}\]
D.\[\dfrac{1}{50}\]

Answer Verified Verified
Hint:Write the sample space of 50 tickets first. Then use both the conditions separately and write their sets. As we have to find the probability of both the conditions combined therefore take the intersection of both the sets and then find the probability by using the formula \[P\left( A\cap B \right)=\dfrac{n\left( A\cap B \right)}{n(S)}\] .

Complete step by step answer:
As we have given the 50 tickets with numbers therefore we can write the sample space as follows,
S = { 00, 01, 02, 03, 04 , 05, 06, 07, 08, 09, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47 48, 49}
Therefore,
 n(s) = 50 …………………………………………………. (1)
Let, A be the set of tickets such that sum of the digits on the selected ticket is 8 therefore we will get,
A = {08, 17, 26, 35, 44}
Therefore, number of elements in A is given by,
n(A) = 5
Also, B be the set of tickets such that the product of its digits is zero therefore we will get,
B = {01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40}
Therefore, number of elements in B is given by,
n(B) = 13
As we have to find the probability such that both the conditions given above must be satisfied and therefore we have to find the set such that its element will presents in both A and B, therefore we will get,
\[\left( A\cap B \right)=\{08\}\]
Therefore number of elements in \[\left( A\cap B \right)\] is given by,
\[n\left( A\cap B \right)=1\] ……………………………………………………… (2)
Now, to proceed further in the solution we should know the formula given below,
Formula:
Probability = P (A) = \[\dfrac{n(A)}{n(S)}\]
Where, A – any set
     S – Sample space.
If we use the above formula and the values of equation () and equation (2) we will get,
\[P\left( A\cap B \right)=\dfrac{n\left( A\cap B \right)}{n(S)}\]
\[\therefore P\left( A\cap B \right)=\dfrac{1}{50}\]
Therefore the probability that the sum of the digits on the selected ticket is 8 and the product of these digits is zero is equal to \[\dfrac{1}{50}\].
Therefore the correct answer is option (d)

Note: In this type of problem always write the sample space if possible which will reduce the silly mistakes. Also do remember that whenever you have to consider two conditions combined you have to take the intersection of both the sets to get the answer.

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