A bag contains 18 balls out of which x balls are white.

A bag contains 18 balls out of which x balls are white.
(i) If one ball is drawn at random from the bag, what is the probability that it is white ball?
(ii) If 2 more white balls are put in the bag, the probability of drawing a white ball will be $\\ \frac { 9 }{ 8 }$ times that of probability of white ball coming in part (i). Find the value of x.

Solution:

Total numbers of balls in a bag = 18
No. of white balls = x
(i) One ball is drawn a random
Probability of being a white ball = $\\ \frac { x }{ 18 }$
(ii) If 2 more white balls an put, then number of white balls = x + 2
and probability is $\\ \frac { 9 }{ 8 }$ times

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test Q13.1

A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is :
(i) a red face card
(ii) neither a club nor a spade
(iii) neither an ace nor a king of red colour
(iv) neither a red card nor a queen
(v) neither a red card nor a black king.

Solution:

Number of cards in a pack of well-shuffled cards = 52
(i) Number of a red face card = 3 + 3 = 6
Probability of red face card will be
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$ = $\\ \frac { 6 }{ 52 }$ = $\\ \frac { 3 }{ 26 }$
(ii) Number of cards which is neither a club nor a spade = 52 – 26 = 26
Probability of card which’ is neither a club nor a spade will be
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$ = $\\ \frac { 26 }{ 52 }$ = $\\ \frac { 1 }{ 2 }$
(iii) Number of cards which is neither an ace nor a king of red colour
= 52 – (4 + 2) = 52 – 6 = 46
Probability of card which is neither ace nor a king of red colour will be
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$ = $\\ \frac { 46 }{ 52 }$ = $\\ \frac { 23 }{ 26 }$
(iv) Number of cards which are neither a red card nor a queen are
= 52 – (26 + 2) = 52 – 28 = 24
Probability of card which is neither red nor a queen will be
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$ = $\\ \frac { 24 }{ 52 }$ = $\\ \frac { 6 }{ 13 }$
(v) Number of cards which are neither red card nor a black king
= 52 – (26 + 2) = 52 – 28 = 24
Probability of cards which is neither red nor a black king will be
P(E) = $\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$ = $\\ \frac { 24 }{ 52 }$ = $\\ \frac { 6 }{ 13 }$