Question

You have a biased dice (with six faces numbered 1,2,3,4,5 and 6) in that even numbers are twice as likely as odd numbers to show. You toss the dice 15 times. What is the probability that

An even number would show up 5 times?

1 would show up five times?

Can it be formulated as a Binomial distribution for answering (a)? If so, how would you solve it?

Can it be formulated as a Binomial distribution for answering (b)? If so, how would you solve it?

Answer #1

Let A: the event that an even number appears

and B: the event that an odd number appears.

Let, P[A]= p , then ac to question, P[B]= p/2;

We know. P[A]+ P[B]= 1

or, p+ p/2 = 1

or, p = 2/3 ;

Now, let X_{i}= 1, if an even number appears in
i^{th} draw, with probability p=2/3

= 0 , if an odd number appears in i^{th} draw, with
probability q= 1-p = 1/3 ; i= 1(1)15 ;

So, Total no of even number ( Y= ) obtained in 15 draws should follow a binomial distribution with parameters n= 15 and p =2/3 .

So, P[Y=5] =

= 0.0067 , which is the probability of obtaining exact 5 even numbers.

Though if you want at least 5 even numbers, the probability would be P[Y>=5]

= 1 - P[Y<5]

=1- P[Y=4]-P[Y=3]-P[Y<2] -P[Y=1]-P[Y=0]

If you have any doubts, let me know in the comment section. Thank you,

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