Probability on cards and bullet hitting target.

probability

(i) The probability of piercing target by a bullet is $\frac{1}{4} $ . Then if $20$ bullets are fired simultaneously , then what is the probability of getting the target?

(ii) What is the probability of not getting king of same colours when $2$ cards are drawn from $52$ with replacement ?

MY WORK:
For (i) , I say that if none of the bullets hit the target, then the probability is :
$$\frac{3^{20}}{4^{20}}$$

Then, probability that one or more bullet hits target :
$$1-\frac{3^{20}}{4^{20}}$$

For (ii ) , I do as follows :

$$\frac{\binom{2}{1}\times \binom{2}{1}}{\binom{52}{1}\times \binom{52}{1}} +\frac{\binom{2}{1}\times \binom{2}{1}}{\binom{52}{1}\times \binom{52}{1}}$$

Then I subtract the above value from $1$ to get the answer … Again , if I take $1$ from red and other from black, the condition holds… Then , probability :

$$\frac{\binom{2}{1}\times \binom{2}{1}}{\binom{52}{1}\times \binom{52}{1}}+ \frac{\binom{2}{1}\times \binom{2}{1}}{\binom{52}{1}\binom{52}{1}}$$

But both the answers seem to clash. Where is the main logic? Which one is correct?

Best Answer

The reasoning for (i) is ok.

For (ii):

There are $|\Omega|=52^2=2704$ total possibilities of drawing two cards from the deck with replacement. So there are two ways of not getting the king of the same color:

  1. As a first card draw non-king, then the second one could be any card $$|A|=\binom{48}{1}\binom{52}{1}=2496$$
  2. As a first card draw a king, then the other card shoud not be a king with the same colour: $$|B|=\binom{4}{1}\binom{50}{1}=200$$

Of course $A\cap B = \emptyset$, so $$P=\frac{|A|+|B|}{|\Omega|}=\frac{2696}{2704}\approx .997$$

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