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Get access to the detailed solutions to the previous years questions asked in IIFT exam
Explanation:
Case 2: When we win by uncovering just 3 spots.
_ _ P _ _ _ _ _ _ _
From the first two uncovered spots 1 will show up P. Out of remaining 7 spots, 3 spots will be filled by No prize. Total number of ways =
2C1*7C3*5!
There are a total of 10 spots out of which 3 are of one type (No Prize), 2 are of one time (The one which will give us prize) and 5 are
different. Therefore, total number of combination in which we can uncover these spots =
We win if we win the two same card before any of the No prize spot. We can win by uncovering just 2 spots and a maximum of 7 spots.
Let 'P' denotes the occurrence of winner card.
Case 1: When we win by uncovering just 2 spots.
P P _ _ _ _ _ _ _ _
Out of remaining 8 spots, 3 spots will be filled by No prize and 5 with different signs. Total number of ways = 8C3*5!
Case 2: When we win by uncovering just 3 spots.
_ _ P _ _ _ _ _ _ _
From the first two uncovered spots 1 will show up P. Out of remaining 7 spots, 3 spots will be filled by No prize. Total number of ways =
2C1*7C3*5!
Case 3: When we win by uncovering just 4 spots.
_ _ _ P _ _ _ _ _ _
From the first three uncovered spots 1 will show up P. Out of remaining 6 spots, 3 spots will be filled by No prize. Total number of ways
= 3C1*6C3*5!
Case 4: When we win by uncovering just 5 spots.
_ _ _ _ P _ _ _ _ _
From the first four uncovered spots 1 will show up P. Out of remaining 5 spots, 3 spots will be filled by No prize. Total number of ways =
4C1*5C3*5!
Case 5: When we win by uncovering just 6 spots.
_ _ _ _ _ P _ _ _ _
From the first five uncovered spots 1 will show up P. Out of remaining 4 spots, 3 spots will be filled by No prize. Total number of ways =
5C1*4C3*5!
Case 6: When we win by uncovering just 7 spots.
_ _ _ _ _ _ P _ _ _
From the first six uncovered spots 1 will show up P. Out of remaining 3 spots, 3 spots will be filled by No prize. Total number of ways =
6C1*3C3*5!
Hence, the probability that a customer will win =
8C3 ∗ 5! + 2C1 ∗ 7C3 ∗ 5! + 3C1 ∗ 6C3 ∗ 5! + 4C1 ∗ 5C3 ∗ 5! + 5C1 ∗ 4C3 ∗ 5! + 6C1 ∗ 3C3 ∗ 5!
3! ∗ 2! ∗ 5!(56 + 70 + 60 + 40 + 20 + 6)
= 10!