Practice Test: Probability - Class 10 MCQ

We know Probability law which is law of  complement
Which is P(not A)=1-P(A)
Where P(A)=⅖
P(not A)=1-⅖=3/5

Given : A die is thrown once .

A die has 6 faces marked as 1, 2, 3, 4, 5 and 6.

If we throw one die then there possible outcomes are as follows: 1, 2, 3, 4, 5 and 6

Number of possible outcomes are = 6

Let E = Event of getting a getting a multiple of 3

Multiples of 3 are = 3, 6

Number of outcome favourable to E = 2

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 2/6  = 1/3

Hence, the probability of getting a  multiple of 3, P(E) = 1/3

Probability of getting a number greater than 2 is
number greater than 2 = 4
total outcomes = 6
so, 4/6
=2/3

Between 1 and 6, meaning not 1 or 6, then we have four favorable choices (2, 3, 4, 5) out of six possible outcomes (1, 2, 3, 4, 5, 6). Therefore your probability is 4/6 = 2/3.

Assume that is an equal chance of the coin landing on heads or tail (the coin is fair, not biased).

Probability of coin landing on head and then on tails=1/2 x 1/2=1/4

Probability of coin landing on tails and then on heads= 1/2 x 1/2 =1/4

Therefore, probability of getting one head and one tail in two coin tosses=1/4+1/4=1/2

Two coins are tossed simultaneously, so there are four possible outcomes ie, HH,HT,TT,TH
Total number of outcomes=4
Probability of getting no head=no. of favourable outcomes/ total no. of outcomes
=1/4

Two coins are simultaneously tossed.
So sample space={HH,HT,TH,TT}
No. of favourable outcomes=getting at least one head={HH,HT,TH}
=3
Total number of outcomes=4
Probability of getting at least one head=No. of favourable outcomes/Total number of outcomes
=3/4

Total number of outcomes = 52
Favourable outcomes = four cards are of Jack.
Total number of favourable outcomes = 4

Probability = Number of favourable outcomes/ Total number of outcomes.

Required probability = P(jack) = 4/52= 1/13.

Hence, the probability of getting Jack is 1/13.

Total Cards: 52

Total Face cards: 12 (4 Jacks, 4 Queens, 4 Kings)

P(Choosing a Face card) =12/52= 3/13

 Let G be the favourable outcomes of getting 10 of black suit then

⇒ n(G) = 2

Therefore,  P(G) 

Let D be the favourable outcomes of getting red and a king, then n(D)=2 Therefore,

Total number of outcomes=7
No. of favourable outcomes=3 (no. Of green balls)
Probability of getting a green ball=No. Of favourable outcomes/Total no. of outcomes=3/7

The probability of an event cannot be negative in any case, thus −1.5 cannot be the probability of an event. The probability of happening of an event always lies between 0 to 1, i.e., 0≤P(E)≤1.

We have one and only even prime number which is 2. So no. of favourable outcomes= 0  and total no. of outcomes=17
Probability of getting 2 =0 this means that 2 cannot occur .

When two coins are tossed, the sample space of possible outcomes would be the set

S = {HH, HT, TH, TT}

where H: the outcome is a “Head” and T: the outcome is a “Tail” and the probabilities associated with each of these outcomes are equal assuming the coins are fair.

When we say at most 1 “Head” outcome, it means that only 1 of the two coins show up a head or that the no “Head” outcome shows up.

If A is the event obtaining only 1 “Head” outcome, then the set of outcomes that makes up event A is

A = {(HT), (TH)}

Hence, the probability of event A is

P(A) = 2/4 

When two coins are tossed simultaneously then the possible outcomes obtained are {HH, HT, TH, and TT}.

Here H denotes head and T denotes tail.

Therefore, a total of 4 outcomes obtained on tossing two coins simultaneously.

Consider the event of obtaining at the most one head. At most one head is obtained means either no head is obtained or one head is obtained.

The outcome(s) favourable to no head obtained is {TT}

The outcome(s) favourable to one head obtained is {HT, TH}

Therefore, the event of obtaining at most one head has 3 favourable outcomes. These are TT, HT and TH.

Therefore, the probability of obtaining at most one head = 3/4.