Question

You have a deck of 52 playing cards.

(i) How many different 8 card hands can be dealt?

(ii) What is the probability that a hand of 8 dealt randomly contains (exactly) 2 aces?

(iii) What is the probability that a hand of 7 dealt randomly will have 7 cards of the same

suit?

An influenza vaccine is produced by two different companies. It is known that a vaccine

produced by company 1 is effective with probability 0:89, while a vaccine produced by

company 2 is effective with probability 0:93. We also know that company 1 supplies 40%

of the vaccines, while company 2 supplies 60% of the vaccines ordered by the government.

(i) What is the probability that a vaccine is effective, given that it was produced by company2?

(ii) What is the probability that a randomly chosen vaccine from the government’s order

is not effective?

(iii) What is the probability that given a vaccine is not effective that it was produced by

company 1?

Answer 1

$a)$

$i)$

We apply combinations to find the 8 card hands can be dealt as follows,

number of ways =$\binom{n}{k}={n!\over(k!(n-k)!)}$

From the question, $n=52, k=8$

Therefore,

$\binom{52}{8}={52!\over(8!\times44!)}=752538194$

Thus, there are 752538194-8card hands that can be dealt.

$ii)$

To get exactly 2 aces, we need to choose 2 of the 4 aces and 6 of the other 48 cards. The number of ways to do that is

$\binom{4}{2}\times\binom{48}{6}={4!\over(2!*2!)}*{48!\over(6!*42!)}=6\times12271512=73629072$

Therefore, the probability is, ${73629072\over752538194}=0.0978(4dp)$

$iii)$

We first choose 7 cards out of a total of 13 cards in a suit. But we only want one out of 4 suits. Therefore, the number of ways for choosing 7 cards out of a total of 13 cards is $\binom{13}{7}={13!\over(7!\times6!)}=1716$ and the number of ways of choosing 1 out out 4 suits is, $\binom{4}{1}={4!\over(1!\times3!)}=4$

Total number of 1716*4=6864

Hence the probability is, ${6864\over752538194}=9.121132e^{-6}$

Therefore, the probability  that a hand of 7 dealt randomly will have 7 cards of the same

suit is =9.121132e^{-6}

$b)$

Let $C_1$ and $C_2$ be the events that the influenza vaccine is manufactured by company 1 and 2 respectively. Also, let $E$ be the event that the influenza vaccine produced is effective. $E'$ is the event that the influenza vaccine produced is not effective.

The following probabilities are given,

$p(E|C_1)=0.89,\space p(E|C_2)=0.93,\space p(C_1)=0.40,\space p(C_2)=0.60$

$i)$

The probability that a vaccine is effective, given that it was produced by company 2 is given as,

$p(E|C_2)=0.93$ as stated above.

$ii)$

We need to determine the probability that a randomly selected vaccine is effective. To do so, we shall apply the law of total probability as follows,

$p(E)=p(E|C_1)*p(C_1)+p(E|C_2)*p(C_2)=0.89*0.40+0.93*0.60=0.356+0.558=0.914$

The probability that a randomly selected vaccine is not effective is given as,

$p(E')=1-p(E)=1-0.914=0.086$

Therefore, the probability that a randomly selected vaccine is not effective is 0.086.

$iii)$

We determine the conditional probability, $p(C_1|E')$ defined as,

$p(C_1|E')={p(C_1\cap E')\over p(E')}$

$P(C_1\cap E')=0.4\times0.11=0.044\space and \space p(E')=0.086$

Thus, $p(C_1|E')={0.044\over 0.086}=0.5116(4dp)$

Therefore, the probability that a randomly selected vaccine is produced by company 1 given it is not effective is 0.5116.