Free Probability 02 Practice Test - 10th Grade

Question 1

If three coins are tossed simultaneously, then the probability of getting at least one head and tail is _____.

\frac{1}{4}

\frac{1}{2}

\frac{3}{4}

\frac{2}{3}

SOLUTION

Solution : C

Given, a coin is tossed 3 times.

Total possible outcomes = {HHH, HHT, HTT, HTH, THH, TTH, THT, TTT} (where H = Heads, T= Tails)
Total no. of possible outcomes = 8
Favourable outcomes (getting at least one head and tail) = {HHT, HTT, HTH, THH, TTH, THT}
No. of favorable outcomes = 6
Probability of an event $E$ ,
$P (E) = \frac{number of favourable outcomes}{total number of outcomes}$

$\Rightarrow$ P (getting at least one head and tail) = $\frac{6}{8}$ = $\frac{3}{4}$
$∴$ The probability of getting at least one head and a tail is $\frac{3}{4}$ .

Question 2

A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is black or king.

\frac{1}{2}

\frac{7}{13}

\frac{15}{26}

\frac{8}{13}

SOLUTION

Solution : B

Total no. of outcomes = 52
Number of black cards (Spade+Club) in a pack of '52' cards = 26
Number of 'Kings' in a pack of cards = 4
Number of 'Black Kings' that have already been included in the number of black cards = 2
$∴$ Number of favourable outcomes $= 26 + 4 - 2 = 28$

$∴$ Probability(getting a black or king) $= \frac{28}{52} = \frac{7}{13}$

Question 3

From a set of 17 cards, numbered 1, 2, ..., 17, one card is drawn at random. What is the probability that number on the drawn card is a multiple of 3 or 7 ?

$\frac{5}{17}$

$\frac{7}{17}$

$\frac{8}{17}$

$\frac{6}{17}$

SOLUTION

Solution : B

The total number of possible outcomes(counting of the cards from 1 to 17) = 17

Favourable outcomes are 3, 6, 7, 9, 12, 14 and 15.
$∴$ No. of favourable outcomes = 7

Let $E$ be the event of getting a multiple of 3 or 7.
$∴$ $Probability P(E)$ = $\frac{Number of outcomes favorable to E}{Number of all possible outcomes of the experiment} = \frac{7}{17}$

Question 4

There are 5 green, 6 black and 7 white balls in a bag. A ball is drawn at random from the bag. Find the probability that it is not white.

$\frac{11}{18}$

$\frac{7}{18}$

$\frac{2}{3}$

$\frac{5}{18}$

SOLUTION

Solution : A

Given,
Number of green balls = 5
Number of black balls = 6
Number of white balls = 7
Total number of outcomes = 5 + 6 + 7 = 18
There are 18 balls out of which 11 are not white.
$\Rightarrow$ Number of favourable outcomes = 11

$Probability of an event, P(E) = \frac{Number of favourable outcomes}{Total number of outcomes}$
$\Rightarrow$ P( ball drawn is not white) = $\frac{11}{18}$

$∴$ Probability that the ball drawn is not white is $\frac{11}{18}$ .

Alternate Method:
P (ball drawn is white) = $\frac{7}{18}$
By complementary event formula,
P( ball drawn is white) + P( ball drawn is not white) = 1
$\Rightarrow$ P( ball drawn is not white)
$= 1 - P( ball drawn is white)$
$= 1 - \frac{7}{18} = \frac{11}{18}$
$∴$ Probability that the ball drawn is not white is $\frac{11}{18}$ .

Question 5

If P(A) and P(not A) are complementary events and P(A) = 0.15, then P(not A) = ?

0.35

0.3

0.85

Cannot be determined

SOLUTION

Solution : C

Given, P (A) = 0.15
As, P(A) and P(not A) are complementary events, P(A) + P(not A) = 1
P (not A) = 1 – P (A) = 1 – 0.15 = 0.85

Question 6

State whether the given statement is true or false:

The probability of getting a multiple of 2 in a throw of an unbiased die is $\frac{1}{2}$ .

True

False

SOLUTION

Solution : A

There are 3 favourable outcomes out of a total of six outcomes in this case.
Multiples of 2 $\leq$ 6 are 2, 4, and 6.
Hence, the probability is $\frac{1}{2}$ .

Question 7

What is the probability of not picking a face card when you draw a card at random from a pack of 52 cards?

$\frac{1}{13}$

$\frac{4}{13}$

$\frac{10}{13}$

$\frac{12}{13}$

SOLUTION

Solution : C

Since there are 12 face cards in a deck of 52cards, the probability of drawing a face card is $\frac{12}{52} = \frac{3}{13}$
Hence, the probability of not picking a face card = $1 - \frac{3}{13} = \frac{10}{13}$

Question 8

A bucket contains 10 brown balls, 8 green balls, and 12 red balls and you pick one randomly without looking. What is the probability that the ball will be brown?

$\frac{4}{15}$

$\frac{1}{3}$

0.61

0.33

SOLUTION

Solution : B and D

There are a total of 10 + 8 + 12 = 30 balls, out of which 10 are brown.
The required probability is $\frac{10}{30}$ = $\frac{1}{3}$ .

Question 9

In a circular dartboard of radius 20 cm, there are 5 concentric circles. the radius of each inner concentric circle is 4 cm less than the outer concentric circle. Find the probability that a dart hits anywhere in the smallest circle assuming that the dart doesn't hit on the boundary of any circle.

$\frac{1}{5}$

$\frac{1}{25}$

$\frac{1}{π}$

SOLUTION

Solution : B

Difference in radius between 2 circles = 4cm

Let, radius of small circle be x.
$\Rightarrow$ x + 4 + 4 + 4 + 4 = 20 (from the diagram)
$\Rightarrow$ x = 4cm

Probability that the dart hits anywhere in the small circle = $\frac{a r e a (i n n e r m o s t c i r c l e)}{a r e a (o u t e r m o s t c i r c l e)}$

$= \frac{π {.4}^{2}}{π 20^{2}}$

$= \frac{1}{25}$

Question 10

The probability of winning a game is $\frac{2}{5}$ , the probability of losing is__. (in decimal form)

SOLUTION

Solution : Winning or losing a game are complementary events. We know that, for complementary events, P(A) + P(notA) = 1.
Thus, if P(winning) = $\frac{2}{5}$
Then, P(losing) = 1 - $\frac{2}{5}$
$\Rightarrow$ P(losing) = $\frac{3}{5}$ = 0.6.

Free Probability 02 Practice Test - 10th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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