We know that there are 52 cards in a pack of cards.

And there are 13 cards of each suit.

Let p be the probability of drawing a card of heart from a pack of 52 cards.

Then,

[∵ there are 13 cards of heart in the pack]

Also, p + q = 1

Where if p is the probability of getting a heart out of 52 cards, then q is the probability of not getting a heart out of 52 cards.

⇒ q = 1 – p

Putting the value of p in the above equation, we get

Now, let the card be drawn n times.

And let X denote the number of hearts drawn out of a pack of 52 cards.

Then, the binomial distribution is given by,

P (X = r) = ⁿC_rp^rq^n-r

Putting and above, we get

…(A)

Where r = 0, 1, 2, 3… n

(i). We need to find the number of times a card can be drawn so that at least there is an even chance of drawing a heart.

In simple words, since n is the number of times a card is drawn, we need to find the smallest n for which P (X = 0) is less than 1/4 satisfies.

So,

From equation A, we have

First, put n = 0.

But 1 ≮ 0.25

Now, put n = 1.

⇒ 0.75 < 0.25

But 0.75 ≮ 0.25

Now, put n = 2.

⇒ 0.5625 < 0.25

But 0.5626 ≮ 0.25

Now, put n = 3.

⇒ 0.42 < 0.25

But 0.42 ≮ 0.25

Now, put n = 4.

⇒ 0.31 < 0.25

But 0.31 ≮ 0.25

Now, put n = 5.

⇒ 0.23 < 0.25

Thus, smallest n = 5.

∴, we must draw cards at least 5 times.

(ii). We need to find the number of times a card must be drawn so that the probability of drawing a heart is more than 3/4.

If P (X = 0) is the probability of not drawing a heart at all.

Then, 1 – P (X = 0) is the probability of drawing a heart out of 52 cards.

According to the question,

First, put n = 0.

But 1 ≮ 0.25

Now, put n = 1.

⇒ 0.75 < 0.25

But 0.75 ≮ 0.25

Now, put n = 2.

⇒ 0.5625 < 0.25

But 0.5626 ≮ 0.25

Now, put n = 3.

⇒ 0.42 < 0.25

But 0.42 ≮ 0.25

Now, put n = 4.

⇒ 0.31 < 0.25

But 0.31 ≮ 0.25

Now, put n = 5.

⇒ 0.23 < 0.25

Thus, smallest n = 5.

∴, we must draw cards at least 5 times.