Answer
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Hint: In the problem, we are asked to find the probability of getting an ace or a spade is asked. Therefore, we need to find the union of both events. The formula for finding the union of two events is $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$
Complete step by step solution:
Let the event of getting an ace card be \[A\] and the event of getting a spade be \[B\].
The probability of occurring of event \[A\] is \[P\left( A \right)\]and similarly, the probability of occurring of event\[B\]is \[P\left( B \right)\].
There are 4 aces in a deck of cards (ace of heart, spade, diamond, club).
Therefore, the probability of the event \[A\] is \[P\left( A \right)=\dfrac{\text{Number of favourable outcome}}{\text{Total number of favourable outcomes}}\]
Therefore, \[P\left( A \right)=\dfrac{4}{52}.............\left( i \right)\]
There are 13 cards of each type in the deck of cards (heart, spade, diamond, club).
Therefore, the probability of the event \[B\] is \[P\left( B \right)=\dfrac{\text{Number of favourable outcome}}{\text{Total number of favourable outcomes}}\]
Therefore, \[P\left( B \right)=\dfrac{13}{52}.............\left( ii \right)\]
There is one card which comes under event \[A\] as well as event $B$. So we need to make sure that we don’t count repeatedly.
Therefore, the intersection of event \[A\] and event \[B\] is \[A\cap B\] = Number of a favourable outcome.
Therefore, \[A\cap B=1\].
The probability of the event \[A\cap B\] is \[P\left( A\cap B \right)=\dfrac{\text{Number of favourable outcome}}{\text{Total number of favourable outcomes}}\].
Therefore, $P\left( A\cap B \right)=\dfrac{1}{52}....................\left( iv \right)$.
By using the formula, $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$
From $\left( i \right),\left( ii \right),\left( iii \right),\left( iv \right)$and substituting the values in the above equation, we get,
$P\left( A\cup B \right)=\dfrac{4}{52}+\dfrac{13}{52}-\dfrac{1}{52}=\dfrac{16}{52}$
Simplifying the equation,
$P\left( A\cup B \right)=\dfrac{16}{52}=\dfrac{4}{13}$
Hence, $P\left( A\cup B \right)=\dfrac{4}{13}$is the required solution and option (a) is correct.
Note: It is important to subtract the intersection term as it will be counted twice. Whenever the probability of finding an event \[A\] or $B$ is asked we need to find the union and whenever the probability of finding an event \[A\] and $B$ is asked we need to find the intersection. The main key in this problem is to remember the formula and be aware of how the deck of cards works.
Complete step by step solution:
Let the event of getting an ace card be \[A\] and the event of getting a spade be \[B\].
The probability of occurring of event \[A\] is \[P\left( A \right)\]and similarly, the probability of occurring of event\[B\]is \[P\left( B \right)\].
There are 4 aces in a deck of cards (ace of heart, spade, diamond, club).
Therefore, the probability of the event \[A\] is \[P\left( A \right)=\dfrac{\text{Number of favourable outcome}}{\text{Total number of favourable outcomes}}\]
Therefore, \[P\left( A \right)=\dfrac{4}{52}.............\left( i \right)\]
There are 13 cards of each type in the deck of cards (heart, spade, diamond, club).
Therefore, the probability of the event \[B\] is \[P\left( B \right)=\dfrac{\text{Number of favourable outcome}}{\text{Total number of favourable outcomes}}\]
Therefore, \[P\left( B \right)=\dfrac{13}{52}.............\left( ii \right)\]
There is one card which comes under event \[A\] as well as event $B$. So we need to make sure that we don’t count repeatedly.
Therefore, the intersection of event \[A\] and event \[B\] is \[A\cap B\] = Number of a favourable outcome.
Therefore, \[A\cap B=1\].
The probability of the event \[A\cap B\] is \[P\left( A\cap B \right)=\dfrac{\text{Number of favourable outcome}}{\text{Total number of favourable outcomes}}\].
Therefore, $P\left( A\cap B \right)=\dfrac{1}{52}....................\left( iv \right)$.
By using the formula, $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$
From $\left( i \right),\left( ii \right),\left( iii \right),\left( iv \right)$and substituting the values in the above equation, we get,
$P\left( A\cup B \right)=\dfrac{4}{52}+\dfrac{13}{52}-\dfrac{1}{52}=\dfrac{16}{52}$
Simplifying the equation,
$P\left( A\cup B \right)=\dfrac{16}{52}=\dfrac{4}{13}$
Hence, $P\left( A\cup B \right)=\dfrac{4}{13}$is the required solution and option (a) is correct.
Note: It is important to subtract the intersection term as it will be counted twice. Whenever the probability of finding an event \[A\] or $B$ is asked we need to find the union and whenever the probability of finding an event \[A\] and $B$ is asked we need to find the intersection. The main key in this problem is to remember the formula and be aware of how the deck of cards works.
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