McDonald’s ran a campaign in which it gave game cards to its customers. These game cards made it possible for customers to win hamburgers, French fries, soft drinks, and other fast-food items, as well as cash prizes. Each card had 10 covered spots that could be uncovered by rubbing them with a coin. Beneath three of these spots were “No Prize” signs. Beneath the other seven spots were names of the prizes, two of which were identical. For, example, one card might have two pictures of a hamburger, one picture of a coke, one of French fries, one of a milk shake, one of a $5, one of $1000, and three “No Prize” signs. For this card the customer could win a hamburger. To win on any card, the customer had to uncover the two matching spots (which showed the potential prize for that card)before uncovering a “No Prize”; any card with a “No Prize” uncovered was automatically void. Assuming that the two matches and the three “No Prize” signs were arranged randomly on the cards, what is the probability of a customer winning?

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 = $$\dfrac{10!}{2!*3!}$$

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 = $$\dfrac{8C3*5!+2C1*7C3*5!+3C1*6C3*5!+4C1*5C3*5!+5C1*4C3*5!+6C1*3C3*5!}{\dfrac{10!}{2!*3!}}$$

$$\Rightarrow$$ $$\dfrac{3!*2!*5!(56+70+60+40+20+6)}{10!}$$

$$\Rightarrow$$ $$\dfrac{1}{10}$$. Therefore, option A is the correct answer.