Question

All the spades are taken out from a pack of cards. From these cards, cards are drawn one by one without replacement till the ace of the spade comes. The probability that the ace comes in the \[{4^{th}}\] draw is
A.\[\dfrac{1}{{13}}\]
B.\[\dfrac{{10}}{{13}}\]
C.\[\dfrac{3}{{13}}\]
D.None of these

Answer 1

Hints: There are 4 suits in decks. Each suit contains 13 cards. We will find the probability of the event not getting an Ace in the first three drawn. Then we will find the probability of getting an ace from the rest card. Then multiply all probabilities to find the required probability.

Formula used:
\[{\rm{Probability}} = \dfrac{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{favourable}}\,\,{\rm{outcomes}}}}{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{total}}\,{\rm{outcomes}}}}\]
The probability of independent events\[A\],\[B\], \[C\], and \[D\]is \[P\left( A \right) \times P\left( B \right) \times P\left( C \right) \times P\left( D \right)\].
Complete step by step solution:
When we draw the first card, then the number of cards present in the suit is 13 cards. In 13 cards, there is one Ace.
The number of favourable outcomes is 1.
The number of total outcomes is 13.
The probability of getting an ace is \[ = \dfrac{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{favourable}}\,\,{\rm{outcomes}}}}{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{total}}\,{\rm{outcomes}}}}\]
                                                                \[ = \dfrac{1}{{13}}\]
The probability formula for not happening an event is \[1 - P(E)\].
The probability of not getting an ace is \[ = 1 - \dfrac{1}{{13}}\]
                                                                     \[ = \dfrac{{12}}{{13}}\].
When we draw the second card, then the number of cards present in the suit is 12 cards as one card is already drawn from the suit. In 12 cards, there is one Ace.
The number of favourable outcomes is 1.
The number of total outcomes is 12.
The probability of getting an ace is \[ = \dfrac{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{favourable}}\,\,{\rm{outcomes}}}}{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{total}}\,{\rm{outcomes}}}}\]
                                                                \[ = \dfrac{1}{{12}}\]
The probability of not getting an ace is \[ = 1 - \dfrac{1}{{12}}\]
                                                                     \[ = \dfrac{{11}}{{12}}\].
When we draw the third card, then the number of cards present in the suit is \[12 - 1 = 11\] cards as one card is already drawn from the suit. In 11 cards, there is one Ace.
The number of favourable outcomes is 1.
The number of total outcomes is 11.
The probability of getting an ace is \[ = \dfrac{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{favourable}}\,\,{\rm{outcomes}}}}{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{total}}\,{\rm{outcomes}}}}\]
                                                                \[ = \dfrac{1}{{11}}\]
The probability of not getting an ace is \[ = 1 - \dfrac{1}{{11}}\]
                                                                     \[ = \dfrac{{10}}{{11}}\].
When we draw the third card, then the number of cards present in the suit is \[11 - 1 = 10\] cards as one card is already drawn from the suit. In 10 cards, there is one Ace.
The number of favourable outcomes is 1.
The number of total outcomes is 10.
The probability of getting an ace is \[ = \dfrac{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{favourable}}\,\,{\rm{outcomes}}}}{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{total}}\,{\rm{outcomes}}}}\]
                                                                \[ = \dfrac{1}{{10}}\].
We know that the probability of independent events is the product of the probability of each event.
The probability that the ace comes in the \[{4^{th}}\] draw is \[\dfrac{{12}}{{13}} \times \dfrac{{11}}{{12}} \times \dfrac{{10}}{{11}} \times \dfrac{1}{{10}}\]
                                                                                             \[ = \dfrac{1}{{13}}\]
Hence option A is correct.
Note: A deck consists of 52cards. In a deck, there are 4 suits. The 4 suits are spades, hearts, diamonds and clubs. Each suit contains 13 different cards. Students get confused with deck and suits. Also, they skip the concept that is after drawing a card then the total number of cards will decrease by 1 for each successive drawing.