Probability

A pack of playing card contains 52 cards, divided into 4 suits. These are – Spades, Clubs, Hearts and Diamonds. Each suit has 13 cards. Among these 13 cards, 9 cards are numbered 2 to 10. The remaining four cards are ace, king, queen and jack. Therefore, each suit contains one ace. Hence, there are total 4 aces in the pack. Out of 4 aces, two can be drawn in ⁴C₂ ways. Here, n(E) = ⁴C₂ = 6. Again, from a pack of 52 cards, 2 cards can be drawn in ⁵²C₂ ways.Therefore, n(S) = ⁵²C₂ = 1326 and the required probability is 6/1326 = 1/221.

Even prime no. = 2

Total outcomes = 6

So, probablity = 1/ 6

Let the total number of marbles be x.

Then, according to question, (x - 104) / x = 1/5

Solving for x, gives, x = 130

No. of blue balls = 130 - 104 = 26

Total cases are: (1) Spade (but not ace) and an ace Probability = (2) Spade ace and an ace P (E) = (3) Ace (not of spade) & a spade P (E) = (4) Ace (of spade) and a spade P (E) = Total probability = 1/26 Option (d) is correct.

Chances that exactly 2 of them will be children = Favourable outcomes/Total outcomes = 10/21. Option 1 is correct.

There are 4 queens and 4 kings in a pack of 52 cards. Probability that it is a king or a queen = 8/52 = 2/13 Option (b) is correct.

3 or 18 can only come if (1, 1, 1) or (6, 6, 6) comes. So, the probability that the sum of the numbers, on the dice, is neither 3 nor 18 = 1 - 2/6 x 6 x 6 = 1 - 1/108 = 107/108 Option 4 is correct.

Total numbers upto 30 divisible by 3 = 10 Total numbers divisible by 5 = 6 Numbers divisible by 3 and 5 = 2 Total numbers divisible by 3 and 5 = 10 + 6 - 2 = 14 So, the probability of getting a number divisible by 3 or 5 = 14/30 = 7/15 Option 1 is correct.

Option 3 is correct.

If a wins, then the probability = 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 x 1/2 x 1/2 +……………..= 1/3 Option 1 is correct.

The probability of drawing a red ball from the first container is 4/7. The probability of drawing a blue ball from the second container is 3/7. Since these are independent events, the probability of both occurring is (4/7) * (3/7) = 12/49.

When tossing two coins, the sample space is {HH, HT, TH, TT}. There are 4 equally likely outcomes. Only one outcome (HH) satisfies the condition of two heads, so the probability is 1/4.