
In a game card bearing numbers 1 to 500, one card is taken and put in a box. Each player selects one card at random and that card is not replaced .If the selected card bears a number which is a perfect square of an even number the player wins prize, Find the probability that
1) The first player wins prize
2) The second player wins the prize, if the first has not won.
Answer
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Hint:We will find the number of events that occur for each of the probability and using the formula of the probability of events we will find the probability of given events.
The number of samples in our problem is 500.
That is \[n(S) = 500\]
Probability of the event is \[\;P\left( E \right) = \dfrac{{n(E)}}{{n(S)}}\]
Here \[n(E)\] is a number of events and \[n(S)\] is a number of samples.
Complete step by step answer:
It is given that in a game cards bearing numbers 1 to 500, one card is taken and put in a box.
Therefore, the total number of possible outcomes is given as 500 that is \[n\left( S \right) = 500\].
1) E= Event first player wins prize
That is nothing but the first player selected a card bears a number which is a perfect square of an even number
That is the event has \[E = \left\{ {4,16,36,64,100,144,196,256,324,400,484} \right\}\]
\[n\left( E \right) = 11\]
Then \[\;P\left( E \right) = \dfrac{{n(E)}}{{n(S)}} = \dfrac{{11}}{{500}}\]
Hence, the probability that the first player wins prize is \[\dfrac{{11}}{{500}}\]
2) The first has not won means one card is already taken but that is not a perfect square of an even number
Since repetition is not allowed we will remove the card from total cards, therefore we get
E’= the second player selected a card bears a number which is a perfect square of an even number
That is
Since the first player has not taken any of the above cards.
Hence, the probability that the second player wins the prize, if the first has not won is \[\dfrac{{11}}{{499}}\].
Note:
In this problem, we found the probability of the required results. We have to focus on collecting the set events because we may go wrong in that sense. Here we have collected the perfect square of an even number. Sometimes we may mistakenly take a number which is not an even number. So, we have to focus on this step.
The number of samples in our problem is 500.
That is \[n(S) = 500\]
Probability of the event is \[\;P\left( E \right) = \dfrac{{n(E)}}{{n(S)}}\]
Here \[n(E)\] is a number of events and \[n(S)\] is a number of samples.
Complete step by step answer:
It is given that in a game cards bearing numbers 1 to 500, one card is taken and put in a box.
Therefore, the total number of possible outcomes is given as 500 that is \[n\left( S \right) = 500\].
1) E= Event first player wins prize
That is nothing but the first player selected a card bears a number which is a perfect square of an even number
That is the event has \[E = \left\{ {4,16,36,64,100,144,196,256,324,400,484} \right\}\]
\[n\left( E \right) = 11\]
Then \[\;P\left( E \right) = \dfrac{{n(E)}}{{n(S)}} = \dfrac{{11}}{{500}}\]
Hence, the probability that the first player wins prize is \[\dfrac{{11}}{{500}}\]
2) The first has not won means one card is already taken but that is not a perfect square of an even number
Since repetition is not allowed we will remove the card from total cards, therefore we get
E’= the second player selected a card bears a number which is a perfect square of an even number
That is
Since the first player has not taken any of the above cards.
Hence, the probability that the second player wins the prize, if the first has not won is \[\dfrac{{11}}{{499}}\].
Note:
In this problem, we found the probability of the required results. We have to focus on collecting the set events because we may go wrong in that sense. Here we have collected the perfect square of an even number. Sometimes we may mistakenly take a number which is not an even number. So, we have to focus on this step.
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