Respuesta :
GIVEN:
You draw one card from a 52-card deck. Then the card is replaced in the deck and the deck is shuffled, and you draw again.
Required;
Find the probability of drawing a nine the first time and a diamond the
second time.
Step-by-step solution;
To solve this math problem, we take note of the following;
A standard 52-card deck contains,
[tex]\begin{gathered} 13\text{ }Hearts \\ \\ 13\text{ }Diamonds \\ \\ 13\text{ }Clubs \\ \\ 13\text{ }Spades \end{gathered}[/tex]Each of the suits (that is, hearts, diamonds, clubs and spades) has a 9 card each which means there are four 9s in the entire 52-card deck.
Hence, the probabilty of drawing a 9 is given as follows;
[tex]\begin{gathered} Probability\text{ }of\text{ }[event]=\frac{number\text{ }of\text{ }required\text{ }outcomes\text{ }}{number\text{ }of\text{ }all\text{ }possible\text{ }outcomes} \\ \\ Probability\text{ }[9]=\frac{4}{52} \\ \\ Probability\text{ }[9]=\frac{1}{13} \end{gathered}[/tex]Note that the card is replaced before the commencement of the next experiment.
This means for the second draw, we have a complete 52-card deck.
Therefore, for the probability of drawing a diamond, note that we have 13 diamonds in all. Hence,
[tex]\begin{gathered} Probability\text{ }[diamond]=\frac{13}{52} \\ \\ Probability\text{ }[diamond]=\frac{1}{4} \end{gathered}[/tex]The probability of event A and event B is the product of probabilities. This means;
[tex]Probability\text{ }of\text{ }A\text{ }and\text{ }Probability\text{ }of\text{ }B=P[A]\text{ }\times\text{ }P[B][/tex]Therefore, the probability of drawing a nine the first time and a diamond the second time is given as;
[tex]\begin{gathered} P[9]\text{ }and\text{ }P[diamond]=\frac{1}{13}\times\frac{1}{4} \\ \\ P[9]\text{ }and\text{ }P[diamond]=\frac{1}{52} \end{gathered}[/tex]ANSWER:
[tex]\frac{1}{52}[/tex]