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LiRa [457]
3 years ago
7

A box contains four blue and eight red balls. Jim and Jack start drawing balls from the box, respectively, one at a time, at ran

dom, and without replacement until a blue ball is drawn. What is the probability that Jack draws the blue ball?
Mathematics
1 answer:
aleksandrvk [35]3 years ago
5 0

Answer:

the probability that Jack draws the blue ball is 0.38539

Step-by-step explanation:

Given the data in the question;

box contains

blue balls = 5

red balls = 8

Jim and Jack start drawing balls from the box, respectively, one at a time, at random, and without replacement

Now,

on the first attempted;

⇒ ( 8/13) × ( 5/12) = 10/39  

on the second attempt;

⇒ ( 8/13 ) ( 7/12) ( 6/11 ) ( 5/10 ) = 14/143

on the third attempt;

⇒ ( 8/13 ) ( 7/12) ( 6/11 ) ( 5/10 ) ( 4/9 ) ( 5/8 ) = 35/1287

on the fourth attempt;

⇒  ( 8/13 ) ( 7/12) ( 6/11 ) ( 5/10 ) ( 4/9 ) ( 3/8 ) ( 2/7 ) ( 5/6 ) = 5/1287

so we add everything;  

⇒ 10/39 + 14/143 + 35/1287 + 5/1287

= 0.38539

Therefore, the probability that Jack draws the blue ball is 0.38539

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Answer:

P(X=0)=(24C0)(0.04)^0 (1-0.04)^{24-0}=0.375413  

P(X=1)=(24C1)(0.04)^1 (1-0.04)^{24-1}=0.  

So then the probability that the batch would be accepeted is:

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Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=24, p=0.04)

On this case p=0.04 represent the probability that a random the tablet would be NOT accepted

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

They accept the batch if there is only one or no tablet that doesn’t meet the test specification, so let's find the individual probabilities for this case:

P(X=0)=(24C0)(0.04)^0 (1-0.04)^{24-0}=0.375413  

P(X=1)=(24C1)(0.04)^1 (1-0.04)^{24-1}=0.  

So then the probability that the batch would be accepeted is:

0.375413+0.375413=0.7508

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