লাল বল= ৫ টি, নীল বল= ১২ টি এবং সবুজ বল= ৮টি 
মোট বল = ৫+১২+৮= ২৫টি 
এদের ভিতর সবুজ বল হওয়ার সম্ভাবনা= ৮/২৫।
সবুজ হওয়ার সম্ভাবনা নাই= ১-৮/২৫=১৭/২৫

In a dice, there are 6 numbers and the number 3 is once

Given number of balls = 3 + 5 + 7 = 15

One ball is drawn randomly = 15C1

probability that it is either pink or red

= \(\frac{7{\mathrm C}_1\;+\;3{\mathrm C}_1}{15{\mathrm C}_1}\;=\;\frac{7\;+\;3}{15}\;=\;\frac{10}{15}\;=\;\frac23\)

Required probability is given by P(E) = \(\frac{\mathrm n(\mathrm E)}{\mathrm n(\mathrm S)}\;=\;\frac1{4{\mathrm C}_2}\;=\;\frac16\)

Total no of ways = (14 – 1)! = 13!

Number of favorable ways = (12 – 1)! = 11!

 

So, required probability = \(\left(\frac{\left(\mathbf{11}\boldsymbol!\boldsymbol\times\mathbf3\boldsymbol!\right)}{\mathbf{13}\boldsymbol!}\right)\)

                                  = \(\frac{39916800\times6}{6227020800}\)

                                  = \(\frac{\mathbf{24}}{\mathbf{625}}\)

Here in this  pack of cards n(S) = 52

Let E = event of getting a queen of the club or a king of the heart

Then, n(E) = 2

P(E) = n(E)/n(S) = 2/52 = 1/26

These two events cannot be disjoint because P(K) + P(L) > 1.

 

P(AꓴB) = P(A) + P(B) - P(AꓵB).

 

An event is disjoint if P(A ꓵ B) = 0. If K and L are disjoint P(K ꓴ L) = 0.8 + 0.6 = 1.4

 

And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

Here n(S) = 6 x 6 = 36

E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}

=> n(E)=20

Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.

As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order).

The overall resultant will remain same.

So final amount with the person will be (in all cases):

= 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27

Hence the final result is: 

64 − 27 = 37

A loss of Rs.37

Here,n = 4(children)

P(girl)= 0.5

P(of atleast one girl)= 1 - P(no girls)

                             = 1 - 0.0625 = 0.9375

P(heart cards)=13/52

P(diamond cards)=13/52

P(heart or diamond)=13/52+13/52=1/2

Given total students in the class = 60
Students who are taking Economics = 24 and
Students who are taking Calculus = 32
Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4
Students who are taking calculus only = 32 - 4 = 28
probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.

Probability of not hitting the ball = 1- 1/4 = \(\frac34\)

Then, the probability that the batsman will hit the ball  = \(1-(\frac34)^{3\;}=\frac{37}{64}\)

P(red cards)=26/52

P(black cards)=26/52

P(red or black cards)=26/52+26/52=1

Total number of elementary events = 52

There are 26 red cards,out of which one red card can be drawn in  \(26C_1\) ways =26.

So,required probability = 26/52 = 1/2

Total number of elementary events = \(20C_2\) ways =190

There are 7 white balls out of which one white can be drawn in ⁷ C₁  ways and one ball from remaining 13 balls can be drawn in \(13C_1\) ways

So,favourable events = \(7C_1\times13C_1\)

Therfore,required probability = \(\frac{7C_1\times13C_1}{20C_2}\) = 91/90

 

Number of applicants = 5
On a day, only 1 leave is approved.
Now favourable events =  1 of 5 applicants is approved
Probability that Laxmi priya's leave is granted = 1/5.

Total numbers in die=6

P(multiple of 2)=1/2

P(multiple of 5)=1/6

P(multiple of 2 or 5)=2/3

Probability of occurrence of an event, 

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

⇒ Probability of getting head in one coin = ½, 

⇒ Probability of not getting head in one coin = 1- ½ = ½, 

Hence, 

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads

= \(\left(\frac12\right)^2\times\left(\frac12\right)^9=\left(\frac12\right)^{11}\)

Total number of chairs = (3 + 5 + 4) = 12.

Let S be the sample space.

Then, n(s)= Number of ways of picking 2 chairs out of 12

= 12×11/2×1 = 66

Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.

=> 8C2 = 8×7/2×1 = 28

Therefore required probability = 28/66 = 14/33.

P(black ball)=3/12

P(red ball)=5/12

P(black or red)=3/12+5/12=2/3

Out of 9 persons,4 can be choosen in \(9C_4\) ways =126.

 

Favourable events for given condition = \(2C_2\times7C_2\) = 21.

 

So,required probability = 21/126 =1/6.

n(S) = \({}^{52}C_2\) = 1326

Let  A = event of getting both red cards

and B = event of getting both queens

then \(\left(A\cap B\right)\) = event of getting two red queens

n(A) = \({}^{26}C_2\) = 325,   n(B) = \({}^4C_2\) = 6

\(n(A\cap B)={}^2C_2=1\)

\(\therefore\;P\left(A\right)=\frac{325}{1326},\;P\left(B\right)\;=\;\frac6{1326}\)

\(P\left(A\cap B\right)=\frac1{1326}\)

 

P ( both red or both queens) = \(P\left(A\cup B\right)\)

= \(P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=\) \(\frac{325}{1326}+\frac1{221}-\frac1{1326}\) = \(\frac{55}{221}\)

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15 
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

S = { 1, 2, 3, 4, 5, 6 } 

=> n(S) = 6

E = { 3, 6}

=> n(E) = 2

Therefore, P(E) = 2/6 =1/3

We know that:

When one card is drawn from a pack of 52 cards

The numbers of possible outcomes n(s) = 52

We know that there are 26 red cards in the pack of 52 cards
⇒ The numbers of favorable outcomes n(E) = 26

Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)

∴ required probability = 26/52 = 1/2.

S = { (1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (2, 1), (2, 2),.........(6, 5), (6, 6) }

=> n(S) = 6 x 6 = 36

E = {(6, 3), (5, 4), (4, 5), (3, 6) }

=> n(E) = 4

Therefore, P(E) = 4/36 = 1/9

Multiples of 3 below 20 are 3, 6, 9, 12, 15, 18 
Multiples of 5 below 20 are 5, 10, 15, 20
Required number of possibilities = 10
Total number of possibilities = 20
Required probability = 10/20 = 1/2.

S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }

=> n(S) = 8

E = { HHH, HHT, HTH, THH }

=> n(E) = 4

P(E) = 4/8 = 1/2

P(red ball) = 5/12

P(white ball) = 4/12

P(red or white ball) = 5/12 + 4/12 = 3/4 = 0.75