Question

Five cards - the ten, jack, queen, king and ace of diamonds are well shuffled with their face downwards. One card is picked up at random.
(i) What is the probability that the card is jack?
(ii) If the king is drawn and put aside, what is the probability that the second card picked up is (a) A queen and (b) A ten?
(a) \[\left( i \right)\dfrac{1}{7},\left( ii \right)\left( a \right)\dfrac{1}{6},\left( b \right)\dfrac{1}{2}\]
(b) \[\left( i \right)\dfrac{1}{6},\left( ii \right)\left( a \right)\dfrac{1}{6},\left( b \right)\dfrac{1}{3}\]
(c) \[\left( i \right)\dfrac{1}{5},\left( ii \right)\left( a \right)\dfrac{1}{4},\left( b \right)\dfrac{1}{4}\]
(d) \[\left( i \right)\dfrac{1}{4},\left( ii \right)\left( a \right)\dfrac{1}{4},\left( b \right)\dfrac{1}{3}\]

Answer 1

Hint: Calculate the probability of getting cards as described in each of the following events. Use the fact that the probability of any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.

Complete step-by-step solution -
We have five cards – the ten, jack, queen, king, and ace of diamonds. We have to find the probability of drawing a card as described in the questions. We will calculate the probability in each case using the formula of probability.
(i) We have to find the probability that the card drawn is a jack.
We have 5 possible outcomes, i.e., ten, jack, king, queen and ace of diamonds.
There is only one jack in the given possible outcomes. So, the number of favourable outcomes is 1.
We know that the probability of any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.
Thus, the probability of getting a jack \[=\dfrac{1}{5}\].
(ii) We will now consider the next case. We have drawn a king and put it aside.
So, now, the number of possible outcomes is 4, i.e., the ten, jack, queen, and ace of diamonds.
(a) We have to calculate the probability of drawing a queen.
There is only one queen in the given set of possible outcomes. So, the number of favourable outcomes is 4.
We know that the probability of any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.
Thus, the probability of getting a queen when a king is already drawn \[=\dfrac{1}{4}\].
(b) We have to calculate the probability of drawing a ten.
There is only one ten in the given set of possible outcomes. So, the number of favourable outcomes is 4.
We know that the probability of any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.
Thus, the probability of getting a ten when a king is already drawn \[=\dfrac{1}{4}\].
Hence, the probability of each of the events is \[\left( i \right)\dfrac{1}{5},\left( ii \right)\left( a \right)\dfrac{1}{4},\left( b \right)\dfrac{1}{4}\], which is option (c).

Note: Probability of any event describes how likely an event is to occur or how likely it is that a proposition is true. The value of the probability of any event always lies in the range \[\left[ 0,1 \right]\] where having 0 probability indicates that the event is impossible to happen while having probability equal to 1 indicates that the event will surely happen. We must remember that the sum of probability of occurrence of some event and probability of non-occurrence of the same event is always 1.