Probability problems?

**a69356** · March 27th 2009, 11:15 PM

Hi,

I am new to probablity topic and learning it. I stuck in solving below questions, so need assistance.Please help me out in finding the way to solve these:-

Q1 - The probability of bomb hitting a bridge is 1/2 and 2 direct hits are required to destroy it.The least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is :

Book Answer is - 7 bombs.

Q2 - Out of a pack of 52 cards one is lost; from the remainder of the pack 2 cards are drawn and are found to be spades.Find the chances that the missing card is a spade?

Book Answer is - 11/50

Q3 - seven white balls and three black balls are randomly placed in a row. Find the probability that the no 2 black balls are placed adjacent to each other ?

Book Answer is - 7/15

Any help would be greatly appreciated.

Thanks,
Ashish

**running-gag** · March 28th 2009, 01:42 AM

Originally Posted by a69356

Hi,

I am new to probablity topic and learning it. I stuck in solving below questions, so need assistance.Please help me out in finding the way to solve these:-

Q1 - The probability of bomb hitting a bridge is 1/2 and 2 direct hits are required to destroy it.The least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is :

Book Answer is - 7 bombs.

Hi

Have a look to this link
Binomial distribution - Wikipedia, the free encyclopedia

Suppose that you throw n bombs. Each of them can only have 2 results (success = hitting the bridge with a probability p = 0.5, or failure with a probability q = 1-p = 0.5) and their results are independent.

The random variable X which counts the number of successes follows a binomial distribution

The number of k successes among the n shots is

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

Here this is a very specific case where p = 1-p
You can simplify the expression : $P(X=k) = \frac{1}{2^n}\:\binom{n}{k}$

The number of 0 success among the n shots is $P(X=0) = \frac{1}{2^n}\:\binom{n}{0} = \frac{1}{2^n}$

The number of 1 success among the n shots is $P(X=1) = \frac{1}{2^n}\:\binom{n}{1} = \frac{n}{2^n}$

The number of successes higher than 2 among the n shots is $P(X\geq2) = 1 - \left(P(X=0) + P(X=1)\right) = 1 - \frac{n+1}{2^n}$

The minimum number of shots to get $P(X\geq2) \geq 0.9$ is 7
because for n=6 $P(X\geq2) = 1 - \frac{7}{2^6} = 0.89$ and for n=7 $P(X\geq2) = 1 - \frac{8}{2^7} = 0.94$

**mr fantastic** · March 28th 2009, 03:30 AM

Originally Posted by a69356

[snip]
Q2 - Out of a pack of 52 cards one is lost; from the remainder of the pack 2 cards are drawn and are found to be spades.Find the chances that the missing card is a spade?

Book Answer is - 11/50

[snip]

Pretend that the lost card is actually the first card drawn. Then the two cards drawn are the second and third cards drawn.

Calculate Pr(1st card drawn is spade | next two cards drawn are spades): $\frac{ \left(\frac{13}{52} \right) \cdot \left(\frac{12}{51} \right) \cdot \left(\frac{11}{50} \right) }{\left(\frac{13}{52} \right) \cdot \left(\frac{12}{51} \right) \cdot \left(\frac{11}{50} \right) + \left(\frac{39}{52} \right) \cdot \left(\frac{13}{51} \right) \cdot \left(\frac{12}{50} \right)}$ .

**a69356** · March 28th 2009, 04:08 AM

Thanks a lot running_gag and Mr fantastic. Above posts are very informative and I learned new stuff.

I tried to solve Q3 mentioned in the first post but I am getting the wrong Answer.

Total number of balls (7 white and 3 black) = 10
They can be arranged among themselves in 10!/(7! * 3!)
Hence sample is n(s) = 60
Considering the three black balls as single and other seven white balls
Total number of balls = 8
Number of ways they are arranged among themselves = 8!/7! = 8
Hence n(e) = 8
p(e) = n(e)/n(s) = 8/60 = 2/15.
Therefore the probability of 3 black balls to be together = 2/15
Hence the probability of not placing all the three black balls together =
1 - 2/15 = 13/15

But the book answer -7/15.

Please correct me where I am wrong.

Thanks,
Ashish

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