Aptitude Discussion

Q. |
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 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 fires, one of a milk shake, one of 5 Dollar, one of 1000 Dollar and three "No Prize" signs. For this card the customer could win a hamburger. To win on any card, the customers 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? |

✔ A. |
0.10 |

✖ B. |
0.15 |

✖ C. |
0.12 |

✖ D. |
0.18 |

**Solution:**

Option(**A**) is correct

As per the question there are 10 cover spots out of which

**(i).** Three spots are there with no prize (identical)**(ii).** Two spots of the same sign (Prize)**(iii).** Five other spots which are distinct

Let three spots of no prize are $(x, x, x)$, two spots of same sign are $b (P, P)$ and five other spots are $(A, B,C, D, E)$.

Total number of cases without restriction:

\(=\dfrac{10!}{3!\times 2!}\)

⇒ Total number of favourable cases happen in the following ways which shows sequence of can covering

**CASE I:** Second uncovering is $P$

P P _ _ _ _ _ _ _ _=\(^8C_3\times 5!\times 1\)

**CASE II:**

_ _ P _ _ _ _ _ _ =\(^7C_3\times 5!\times 2\)

**CASE III:**

\(=^6C_3\times 5!\times 3\)

**CASE IV:**

\(=^5C_3\times 5!\times 4\)

**CASE V:**

\(=^4C_3\times 5!\times 5\)

**CASE VI:**

\(=^3C_3\times 5!\times 6\)

Total number of favourable cases:

\(=\dfrac{\text{case (I+II+III+IV+V+VI)}}{10!/3!\times 2!}\)

\(=0.10\)

**Tanner**

*()
*

The much easier way to do this is to acknowledge the 5 distinct places are irrelevant and then to write possible events in the form PXXPX, where P is picking 1/2 of the identical prizes, and X is a no prize.

Only 1 of these events yields a winning card, PPXXX.

There are (5 choose 2) = 10 possible distinct events.

Hence, the probability is 1/10.