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Is it possible to help me to verify my answer to the questions below?

  1. If you shuffle a standard deck of $52$ cards, what is the probability that the top $2$ cards are both "5"?

ANS: $\frac 4{52} \times \frac 3{51} = \frac1{221}$

  1. You roll $2$ dice. You get $\$100$ if at least one of them is $6$, or sum of them is $6$. How much do you think this game is worth?

ANS: $(\frac{11}{36} + \frac5{36}) \times 100 = \frac 49\times 100 = \frac{400}9\approx \$44.44$

  1. A train is approaching a bridge. You are at the $\frac14$ position of the bridge. Running both ways can rescue you (both ways can get you safe, and just before hit by the train). What is the speed ratio between you and the train?

ANS: Suppose the man has speed of $1$ m/s, train is $x$ m/s,

$\frac{x+4}3=\frac x1$

$x=2$

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    Why do you refer to this as "*mental math*"... are you not allowing the use of pen&paper to calculate anything? Why do you answer the first with a ratio and not the second? Question 3 seems badly worded as you could get off the bridge long before the train arrives, so there seems to be no upper limit on the speed of the man. – 2017-02-28
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    @JMoravitz, Question #3 assumes that you can just get it off (either running backward or forward) just before you get hit by the train. – 2017-03-03

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  1. Your first answer is correct.

  2. At least one 6. Means 6 on one die or 6 on both dice.

Probability = $\frac {1}{6} \times \frac {5}{6} + \frac {5}{6} \times \frac {1}{6} + \frac {1}{6} \times \frac {1}{6}$

$ = \frac {5}{36} + \frac {5}{36} + \frac {1}{36}$

$ = \frac {5+5+1}{36} = \frac {11}{36}$

Sum of numbers is 6. Cases are (1,5) , (2,4), (3,3), (4,2), (5,1).

Probability = $\frac {5}{36}$

Total probability = $\frac {11}{36} \times \frac {5}{36}$

So your answer 2 is also correct.

  1. Question is not enough clear. We have many doubts like how apart train when man start running.
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    I guess how aprt train when man start running can be calculated using the problem statement above. Why don't you give it a try? – 2017-03-03