Question

Six cards – ace, jack, queen, king, ten and nine of diamond are well shuffled with their face downwards. One card is then picked up at random.
(i)What is the probability that the card is the jack?
(ii)If the ace is drawn and put aside, what is the probability that the second card picked up is (a) a queen (b) an ace

Answer 1

Hint: Here we will use the concept of the probability to find the probabilities of the different conditions. Probability is the ratio of the possible outcome to the total number of outcomes. The probability of occurrence always lies between 0 to 1.

Complete step-by-step answer:
It is given that the total of six cards – ace, jack, queen, king, ten and nine of diamond are well shuffled with their face downwards. Therefore, we get
Number of cards \[ = 6\]
Now we will find the probability that the card is the jack which is equal to the ratio of the number of jacks to the total number of cards.
As there are only 6 cards and jack has appeared only once, therefore, we get
Probability that the card is the jack \[ = \dfrac{1}{6}\].
Now it is given that if the ace is drawn and put aside. Now the probability that the second card picked up is a queen.
One card is already picked so we are left with only 5 cards. In these 5 cards there is only 1 ace, therefore, we get
Probability that the second card picked up is a queen \[ = \dfrac{1}{5}\]
Now we will find the probability that the second card picked up is an ace which will be equal to zero as there is only one ace in the total cards and the card of ace is already drawn in the first turn. Therefore, we get
Probability that the second card picked up is an ace \[ = 0\] .

Note: Probability is the subject which helps us a lot in the prediction of the events. We use this probability concept in our day-to-day life also like in weather forecasting etc.
Here we have to keep in mind that if the probability of the event is one then that event is known as the true event. We should also note that the sum of the probability of occurrence of an event and sum of the probability of not occurrence of an event is always equal to one.
\[P\](Occurrence of event) \[ + P\]( not occurrence of event) \[ = 1\]