Maths, probability (1 Viewer)

excellence1 · Dec 6, 2015

Paul has 10 playing cards in her hand, consisting of 5 different pairs of cards. She shuffles the cards and places 2 of them face up on th e table
a) Find teh probabilitt that these two cards are not pair
b) A third card is placed on the table. fIND THE probability that ths card forms a pair with one of these two cards already on the table

InteGrand · Dec 6, 2015

excellence1 said:
Paul has 10 playing cards in her hand, consisting of 5 different pairs of cards. She shuffles the cards and places 2 of them face up on th e table
a) Find teh probabilitt that these two cards are not pair
b) A third card is placed on the table. fIND THE probability that ths card forms a pair with one of these two cards already on the table

$\noindent Note that a ``pair'' is two cards of the same rank (e.g. two Queens or two Aces etc.)$

$\noindent (a) The first card on the table can be anything. For the second card, it is one of the remaining nine. Exactly one of these remaining nine cards would form the pair with the first card on the table. Hence the probability the two cards on the table \emph{are} a pair is $\frac{1}{9}$. Hence the probability that they are \emph{not} a pair is $1-\frac{1}{9}=\frac{8}{9}$.$

$\noindent (b) If the two cards initially on the table are already a pair $\Big{(}$which happens with probability $\frac{1}{9}\Big{)}$, then there is 0 chance that the third card can form a pair, since we began with 5 different pairs. If the two cards initially on the table are \emph{not} a pair $\Big{(}$which happens with probability $\frac{8}{9}\Big{)}$, then for the third card to form a pair with one of the other two, we must pick the corresponding card to one of the two on the table initially. There are therefore two cards that we could choose that would result in a pair being formed. Since we choose the third card from a total of eight, this probability is $\frac{2}{8}$.$

$So the required answer is$

$\begin{align*}\frac{1}{9}\cdot 0 + \frac{8}{9} \cdot \frac{2}{8}= \frac{2}{9}.\end{align*}$

leehuan · Dec 6, 2015

It can sometimes help to note that

P(Event occurs) = Favourable outcomes / Total outcomes

And always complementary probabilities:

P(Event doesn't occur) = 1 - P(Event occurs)

Maths, probability (1 Viewer)

excellence1

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InteGrand

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leehuan

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