# All the jacks, queensapd kings are removed

Question:

All the jacks, queensapd kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at

random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value.

(i) 7

(ii) greater than 7

(iii) Less than 7

Solution:

In out of 52 playing cards, 4 jacks, 4 queens and 4 kings are removed, then the remaining

cards are left, n(S) = 52 – 3 x 4 = 40.

(i) Let $E_{1}=$ Event of getting a card whose value is 7

$E=$ Card value 7 may be of a spade, a diamond, a club or a heart

$\therefore \quad n\left(E_{1}\right)=4$

$\therefore \quad P\left(E_{1}\right)=\frac{n\left(E_{1}\right)}{n(S)}=\frac{4}{40}=\frac{1}{10}$

(ii) Let $E_{2}=$ Event of getting a card whose value is greater than 7

$=$ Event of getting a card whose value is 8,9 or 10

$\therefore \quad n\left(E_{2}\right)=3 \times 4=12$

$P\left(E_{2}\right)=\frac{n\left(E_{2}\right)}{n(S)}=\frac{12}{40}=\frac{3}{10}$

(iii) Let $E=$ Event of getting a card whose value is less than 7

$=$ Event of getting a card whose value is $1,2,3,4,5$ or 6

$\therefore \quad n\left(E_{3}\right)=6 \times 4=24$

$\therefore \quad P\left(E_{3}\right)=\frac{n\left(E_{3}\right)}{n(S)}=\frac{24}{40}=\frac{3}{5}$