Option 1 : \(\frac{2}{{91}}\)

__Explanation:__

The total number of out comes in selecting 3 balls out of total 15 balls is ^{15}C_{3}

Total out comes in selecting 3 red balls from total 5 red balls is ^{5}C_{3}

∴ The probability of getting all three red balls is

\(P = \frac{{{5_C}_3}}{{{{15}_{{C_3}}}}} = \frac{{5 \times 4 \times 3}}{{15 \times 14 \times 13}} = \frac{2}{{91}}\)

Option 1 : Positively skewed

**Given**

Mean = 32

Median = 30

**Formula**

Mode = 3 × median - 2 × mean

**Calculation**

Mode = 3 × 30 - 2 × 32

∴ Mode = 26

Here Mode is least and median is lies between mean and mode

**∴ Mean > Median > Mode**

**∴ This relation satisfied Positive skewed **

__Key Points__

Relation of mean, mode, and median for positive skewness is Mean > Median > Mode

Relation of mean, mode, and median for negative skewness is Mode > Median > Mean

**Concept:**

Mean = \(\sum\limits_x {x.{f_x}} \)

**Calculation:**

Value on the die (x) |
1 |
2 |
3 |
4 |
5 |
6 |

Probability P(X) |
1/6 |
1/6 |
1/6 |
1/6 |
1/6 |
1/6 |

Mean = \(\sum\limits_x {x.{f_x}} \)

= 1.(1/6) + 2.(1/6) + 3.(1/6) + 4.(1/6) + 5.(1/6) + 6.(1/6)

= 3.5

Option 1 : Mean

**Explanation**

The mean is the most popular and well-known measure of central tendency. It can be used with both discrete and continuous data. Although its use is most often

with continuous data. The mean is equal to the sum of all the values in the data set divided b the number of values in the data set

Mean = ∑x_{i}/N

N = number of observation

X = observation

**Median** = Median locates a central part which divides a distribution into two equal halves

**Median** = (N + 1/2) th item, N = number of observation

**Mode** = Mode refers that value in a distribution which occur most frequently

**Mode** = M_{o} = L + [(F_{1} - F_{O})/(2F_{1} - F_{O} - F_{2})] \TIMES C

C = Size of the interval

F_{1} = Frequency of the modal class

F_{2} = Frequency of the class succeeding the modal class

F_{0} = Frequency of the class preceding the modal class

L = Lower limit of the modal class

Option 4 : \(\dfrac{1}{6}\)

A committee of 4 is to be formed from among 4 girls and 5 boys.

The probability that the committee will have the number of boys less than the number of girls is,

= The probability of selecting 4 girls + The probability of selecting 3 girls and 1 boy

The probability of selecting 4 girls \( = \frac{{{4_{{c_4}}}}}{{{9_{{c_4}}}}} = \frac{1}{{126}}\)

The probability of selecting 3 girls and 2 boys \( = \frac{{{4_{{c_3}}} \times {5_{{c_1}}}}}{{{9_{{c_4}}}}} = \frac{{20}}{{126}}\)

Now, the required probability \( = \frac{1}{{126}} + \frac{{20}}{{126}} = \frac{{21}}{{126}} = \frac{1}{6}\)

Option 1 : 4/15

A box has 8 red balls and 8 green balls. Two balls are drawn randomly in succession from the box without replacement.

Let P_{1}(E) denotes the probability that the first ball drawn is red.

Let P_{2}(E) denotes the probability that the second ball drawn is green.

As there are total 16 balls and we need to select one red ball

There are 8 red balls.

\({P_1}\left( E \right) = \frac{{{8_{{c_1}}}}}{{16{c_1}}} = \frac{8}{{16}} = \frac{1}{2}\)

Now the total balls will become 15 and the number of green balls is 8.

We need to select a ball from 8 green balls.

\({P_2}\left( E \right) = \frac{{{8_{{c_1}}}}}{{16{c_1}}} = \frac{8}{{15}}\)

As P_{1}(E) and P_{2}(E) both are independent

P(E) = P_{1}(E). P_{2}(E)

A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is

Option 3 : 1/1260

**Explanation:**

The given method is drawing without replacement, therefore each time a thing gets out, the total count decrease by one.

**Given:**

2 Washers, 3 Nuts, 4 bolts = Total of 9 things.

Order of taking out is: 2 washers first followed by 3 nuts and subsequently the 4 bolts

\(P\left( R \right) = \frac{2}{9} \times \frac{1}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} \times \frac{4}{4}\times\frac{3}{3}\times\frac{2}{2}\times \frac{1}{1} = \frac{1}{{1260}}\)

Option 2 : 0.3

**Given**

P(A) = 2P(B) = 6P(C)

**Concept**

When event are mutually exclusive and exhaustive

P(A) + P(B) + P(C) = 1

**Calculation**

Let P(A) be k

⇒ P(B) = k/2

⇒ P(C) = k/6

So according to the concept

⇒ k + k/2 + k/6 = 1

⇒ 10k/6 = 1

⇒ k = 3/5

Option 1 : 2 : 1

**Concept:**

**Odds in favor:**

**\(P(\rm{Odds\;in\;Favor})=\frac{Number\;of\;favourable\;outcome}{No\;of\;unfavorable\;outcome}\)**

Odds in Against:

\(P(\rm{Odds\;Against})=\frac{Number\;of\;unfavourable\;outcome}{No\;of\;favorable\;outcome}\)

**Calculation:**

**Given:**

The probability of occurrence A is = \(\frac{1}{3}\)

The probability of not-occurrence A is = \(\frac{2}{3}\)

Odd against the happening of A is:

\(P(\rm{Odds\;Against})=\frac{Number\;of\;unfavourable\;outcome}{No\;of\;favorable\;outcome}\)

\(P(\rm{Odds\;Against})=\frac{\frac{2}{3}}{\frac{1}{3}}=\frac{2}{1}\)

Option 3 : 140/429

Probability of latter box contains 2 red and 6 green balls \(= \frac{{{5_{{C_2}}} \times {{10}_{{C_6}}}}}{{{{15}_{{C_8}}}}} = \frac{{10 \times 210}}{{6435}} = \frac{{140}}{{429}}\)

Option 2 : 0

For ‘n’ number of coin tosses

for \(n\) head, 0 tails; \(n - 1\) heads ,1 tails; \(n-2\) heads, 2 tails; \(n-3\) heads,3 tails; …………..; 1 heads, \(n - 1\) tails; 0 heads, \(n\) tails.

The difference between the number of heads and tails is \(n,\;n - 2,\;n - 4,\; \ldots ..,\;2,\;0.\)

It is not \(\left( {n - 1} \right)\), \(\;\left( {n - 3} \right),\;\left( {n - 5} \right),\; \ldots \ldots \;,\) etc.

∴ The probability of difference being ‘n – 3’ = 0.Option 1 : P(A) - P(A)P(B)

**Given**

A and B are independent event

P(B /A) = P(B) then A and B are independent event

**Explanation**

P(A – B) = P(A) – P(B)

⇒ P(A – B) = P(A) – P(A ∩ B)

⇒ P(A – B) = P(A) – P(A)P(B) [P(A ∩ B) = P(A).P(B)]

**∴**** P(A – B) is P(A) – P(A)P(B)**

Option 1 : skewness

**Explaination**

We all know that a measure of skewness is obtained by the use of the third moment about the mean (Third central mean)

Skewness is measured by the term β_{1}

⇒ μ^{2}_{3}/μ^{2}_{2}

The first moment about the origin is the measure of the mean

The second moment about the mean measures the variance

The third moment about the mean measures **the skewness **

**Concept:**

As per **Binomial Probability Distribution,**

**P (x = r success) = ^{n}C_{r} (p)^{r} (q)^{n-r}**

Where, n = Total no. of trials

r = number of successful trials

p = probability of success in 1 trial

q = probability of failure is 1 trial = (1 - p)

**Calculation:**

Given,

Number of times coin tossed (n) = 15

Probability of getting head in 1 trial

\(p = \frac{1}{2}\)

\(\therefore q = 1 - \frac{1}{2} = \frac{1}{2}\)

∴ Probability of getting 8 heads exactly,

p (x = 8) =Option 2 : 14 / 45

**Given**

The probability of getting plumbing contract = P(A) = 2/3

Probability of getting not electric contract P(B̅) = 5/9

Probability of getting atleast one contract P(A U B) = 4/5

**Calculation**

The probability of not getting electric contract = 1 - P(B̅)

⇒ = 1- 5/9 = 4/9

Probability of getting both the contracts P(A ∩ B) = P(A) + P(B) - P(A U B)

⇒ P(A ∩ B) = 2/3 + 4/9 - 4/5

⇒ (90 + 60 - 108)/135

**∴ The probability that he will get both the contracts P(A ∩ B) is 14/45**

Option 4 : \(\dfrac{(-480)^2}{(132)^3}\)

**Given**

**μ' _{1}, μ_{2}, μ'_{3}, μ'_{4} are row moments**

μ’_{1} = 2

μ’_{2} = 136

μ’_{3} = 320

μ’_{4} = 40,000

**Calculation**

We have to find the first our central moments

μ_{1} = μ’_{1} = 2

μ_{2} = μ’_{2} – (μ’_{1})^{2}

⇒ 136 – (2)^{2}

_{⇒} 136 – 4

⇒ 132

μ_{3} = μ’_{3} – 3 × μ’_{2}μ’_{1} + 2 × (μ’_{1})^{3}

⇒ 320 – 3 × 136 × 2 + 2 × (2)^{3}

⇒ 320 – 816 + 16

⇒ - 480

μ_{4} = μ’_{4} – 4 × μ’_{1} × μ’_{3} + 6 × μ’_{2} × (μ’_{2})^{2} – 3 × (μ’_{1})^{4}

⇒ 40000 – 4 × 2 × 320 + 6 × 136 × 2^{2} – 3 × 2^{4}

⇒ 40000 – 2560 + 3264 – 48

⇒ 40656

The coefficient of skewness denoted by β_{1}

⇒ β_{1} = μ^{2}_{3}/μ^{3}_{2}

⇒ (-480)^{2}/(132)^{3}

**∴ The coefficient of skewness is (-480) ^{2}/(132)^{3}**

Option 3 : 1/11

**Given**

P(A) = 3/4

P(B) = 9/20

P(A ∩ B) = 1/4

**Calculaton**

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ 3/4 + 9/20 – 1/4

⇒ 19/20

P(A/B) = P(A^{'} ∩ B^{1})/P(B^{'}) = 1 – P(A ∪ B)^{'}/[1 – P(B)]

⇒ [1 – 19/20]/[1 – 9/20]

⇒ 1/11

**⇒ Probability P(A/B) IS 1/11**

Option 1 : 1/8

**Given**

P(A) = 2/3

P(B) = 5/8

P(Only A occurring) = 1/6

**Calculation**

P(A ∩ B) = P(A) – P(Only A occurring)

⇒ 2/3 – 1/6

P(A ∩ B) = ½

Probability of only B occurring = P(B) - P(A ∩ B)

⇒ 5/8 – 1/2

Option 1 : Negatively skewed.

**Concept:**

**Calculation:**

__Given:__

Mean = 40

Mode = 41

Median = 40.5

In negative skewed = Mean is minimum while mode is maximum and median is between mean and mode.

Here Mean < Median < Mode.

**∴ The distribution is negatively skewed.**

** Important Points**:

**In normal, **Mean = Median = Mode.

**In positive skewed,** Mean > Median > Mode.

Option 3 : \(\dfrac{4}{13}\)

**Concept:**

A, B, and C are mutually exclusive events,

P(A) + P(B) + P(C) = 1

**Calculation:**

P(B) = \(\rm \frac{3}{2}P(A)\) and P(C) = \(\rm \frac{1}{2}P(B)\)

⇒ Let P(A) = p

⇒ P(B) = \(\rm \frac{3p}{2}\) and P(C) = \(\rm \frac{3p}{4}\)

Since A, B, C are mutually exclusive and exhaustive events associated with a random experiment.

⇒ A ∪ B ∪ C = S

⇒ P(A ∪ B ∪ C) = P(S)

⇒ P(A ∪ B ∪ C) = 1

⇒ P(A) + P(B) + P(C) = 1

⇒ p + \(\rm \frac{3p}{2}\) + \(\rm \frac{3p}{4}\) = 1

⇒ p = \(\rm \dfrac{4}{13}\)

∴ P(A) = \(\rm \dfrac{4}{13}\)