What Is The Probability That Nick Draws A Queen And Rolls A Sum Of 10?

FACTS  AND  FORMULAE  FOR  PROBABILITY  QUESTIONS

1. Experiment :An operation which can produce some well-defined outcomes is called an experiment.

2. Random Experiment :An experiment in which all possible outcomes are know and the verbal output cannot be predicted in advance, is called a random experiment.

Ex :

i.Tossing a fair coin.

ii. Rolling an unbiased die.

three.Drawing a card from a pack of well-shuffled cards.

3. Details of above experiments:

i. When we throw a coin, then either a Head (H) or a Tail (T) appears.

ii. A dice is a solid cube, having vi faces, marked 1, ii, 3, 4, five, 6 respectively. When we throw a die, the outcome is the number that appears on its upper face.

iii. A pack of cards has 52 cards.

• Information technology has 13 cards of each arrange, name Spades, Clubs, Hearts and Diamonds.
• Cards of spades and clubs are black cards.
• Cards of hearts and diamonds are red cards.

In that location are 4 honours of each unit of measurement. There are Kings, Queens and Jacks. These are all called confront cards.

4. Sample Space:When we perform an experiment, and so the set S of all possible outcomes is called the sample space.

Ex :

1. In tossing a money, Southward = {H, T}

ii. If two coins are tossed, the South = {HH, HT, TH, TT}.

3. In rolling a dice, we have, Southward = {1, 2, 3, 4, 5, 6}.

Effect :Any subset of a sample infinite is called an issue.

v. Probability of Occurrence of an Event :

Let Due south be the sample and let Eastward be an issue.

So,$E\subseteq S$

$\therefore P\left(\mathrm{Eastward}\right)=\frac{n\left(\mathrm{Due\; east}\right)}{n\left(S\right)}$

half dozen. Results on Probability :

i. P(S) = 12.$0\le P\left(E\right)\le 1$iii.$P(\varnothing )=0$

iv. For any events A and B we have :

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

5. If$\overline{A}$ denotes (non-A), and so$P\left(\overline{A}\right)=1-P\left(A\right)$

Q:

The 2nd statement is the ____ of the start.

$$\begin{gathered} 1 . P ⇒ Q \\ 2 . ~ Q ⇒ ~ P \end{gathered}$$

A) Converse	B) Changed
C) Contrapositive	D) Contradiction

A) Converse

B) Inverse

C) Contrapositive

D) Contradiction

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Q:

Tickets numbered one to xx are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 15 ?

A) half-dozen/nineteen	B) 3/10
C) vii/ten	D) 6/17

A) 6/19

B) 3/ten

C) 7/ten

D) six/17

Reply & Explanation Answer: B) 3/10

Explanation:

Hither, Due south = {1, 2, 3, 4, ...., xix, xx}=> n(southward) = 20
Let East = event of getting a multiple of 4 or 15
=multiples od 4 are {iv, 8, 12, 16, twenty}
And multiples of 15 ways multiples of iii and 5
= {3, six , ix, 12, xv, xviii, 5, ten, xv, 20}.
= the mutual multiple is only (xv).
=> East = north(E)= 6
Required Probability = P(E) = n(Eastward)/n(S) = 6/20 = iii/10.

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Q:

The probability that a bowler bowled a brawl from a bespeak will hit by the batsman is ¼. Three such balls are bowled simultaneously towards the batsman from that very bespeak. What is the probability that the batsman will striking the ball ?

A) 37/64	B) 27/56
C) 11/xiii	D) 9/8

A) 37/64

B) 27/56

C) 11/13

D) 9/8

Answer & Explanation Answer: A) 37/64

Explanation:

Probability of not hit the brawl = 1- 1/4 =
Then, the probability that the batsman will hit the ball =

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Q:

Variance of the information ii, 4, five, 6, viii, 17 is 23.33. Then the variance of 4, 8, 10, 12, sixteen, 34 volition be

A) xi.66	B) 46.66
C) 93.3333	D) iv.83

A) xi.66

B) 46.66

C) 93.3333

D) iv.83

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Q:

A group of v persons is formed from five boys and four girls. Find the probability that there are at least ii girls in the group?

A) 110/129	B) 5/6
C) 121/126	D) 3/7

A) 110/129

B) 5/6

C) 121/126

D) iii/7

Answer & Explanation Answer: A) 110/129

Explanation:

Full number of possible ways =

9 C v = ix x eight x 7 x vi x 5 5 ten 4 x 3 ten 2 10 1 = 126

Number of favorable cases =

$$\begin{gathered} 4 C two x 5 C 3 + 4 C 3 10 5 C two + 4 C 4 x 5 C 1 \\ four ten iii 2 x 1 x 5 x 4 10 3 3 x 2 x 1 + 4 ten 3 ten 2 3 x ii x i x 5 10 4 2 ten 1 + 1 x 5 \\ = half dozen 10 x + iv x 10 + five \\ = threescore + 40 + v \\ = 105 \end{gathered}$$

Therefore, required probability = 105/126 = 5/6

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Q:

There are 26 balls marked with alphabetical order A to Z. What is the probability of selecting vowels listed balls?

A) 1	B) 21/26
C) v/26	D) 5

A) i

B) 21/26

C) 5/26

D) 5

Respond & Caption Respond: C) five/26

Explanation:

We know that,

Total number of balls due north(S) = 26

Number of vowels north(E) = v

Hence, required probability = north(E)/northward(S) = v/26.

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Q:

What is the probability of getting a sum 9 from ii throws of a dice ?

A) 1/half-dozen	B) 1/2
C) i/9	D) 3/4

A) 1/6

B) i/2

C) 1/nine

D) 3/4

Respond & Explanation Answer: C) 1/ix

Caption:

In two throws of a die, n(s)=(six x 6)=36

let E= Upshot of geting a sum 9={(iii,6),(four,5),(5,4),(half-dozen,3)}

P(East) = n(E)/n(Southward) = iv/36 = 1/9

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Q:

In the usual fix notation,A ∪ ( B ∩ C ) =

A) A∪B∪A∪C	B) A∩B∪A∩C
C) A∪B∩A∪C	D) A∪B∩A∩C

A) A∪B∪A∪C

B) A∩B∪A∩C

C) A∪B∩A∪C

D) A∪B∩A∩C

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