Question

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

Answer 1

Explanation:

We have n(s) =52C2 52 = 52*51/2*1= 1326.

Let A = event of getting both black cards

   B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = 52*5/12*1 = 26C2 = 325, n(B)= 26*25/2*1= 4*3/2*1= 6  and  n(A∩B)

= 4C2 = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

Answer 2

Answer:

$\frac{55}{221}$

Step-by-step explanation:

Given:

Two cards is drawn from 52 cards

To find:

Probability of either both are black or both are queen

Solution:

We only need to use the fundamental principles of probability in these kinds of queries while keeping in mind that we are not considering the same card twice. Here, the only metric we've utilised is the proportion of favourable results to all outcomes.

In the field of mathematics known as probability, random events are analysed. A random event's outcome cannot be predicted before it happens, although it could take any of several different forms. The final result is thought to have been determined by chance.

52 cards can be used to draw two cards in $^5^2C_2$

n(S)=$^5^2C_2$

Similarly, there are 26 red cards and 26 black cards in the 52-card deck. If both of the cards are black, let A be the outcome.

n(A)=$^2^6C_2$

$P(A)=\frac{n(A)}{n(S)} =\frac{^2^6C_2}{^5^2C_2} =\frac{26\times25}{52\times51} =\frac{25}{102}$

If both cards are queens, then let B be the scenario. The 52 cards in a deck contain 4 queens.

Therefore, there are $^4C_2$ ways to draw two queen cards.

n(B)=$^4C_2$

$P(B)=\frac{n(B)}{n(S)} =\frac{^4C_2}{^5^2C_2}=\frac{4\times3}{52\times51}=\frac{1}{221}$

There are two queens in black hence

$n(A\cap B)=\frac{A\cap B}{n(S)}=\frac{1}{^5^2C2} =\frac{1\times2\times1}{52\times51}$

$=\frac{1}{1326}$

Thus the required probability = $P(A\cup B)$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)\\=\frac{25}{102} +\frac{1}{221} -\frac{1}{1326} \\=\frac{325}{1326} +\frac{6}{1326} -\frac{1}{1326} \\=\frac{330}{1326} \\=\frac{55}{221}$

From a deck of cards, 10 cards are picked at random and shuffled. The cards are as follows: 6, 5, 3, 9, 7, 6, 4, 2, 8, 2 Find the probability of picking a card having value more than 5 and find the probability of picking a card with an even number on it.

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