Probability questions

P(getting head when a coin is tossed)=\(\frac{1}{2}\)

P(getting a 3 when a die is rolled)=\(\frac{1}{6}\)

P(both the above events occure at the same time)=\(\frac{1}{2}\times\frac{1}{6}\)=\(\frac{1}{12}\)

n(S) = 52C3 = 132600/6 = 22100
n(E) = 4C3 = 24/6 = 4
p = 4/22100 = 1/5525

n(S) = 52C3 = 132600/6 = 22100
n(E) = 4C3 = 24/6 = 4
p = 4/22100 = 1/5525

The possibilities are, 3 boys and 2 girls, 4 boys and 1 girl, 5 boys.

Required Probability\( = \frac{9C_{3}\times 5C_{2}+9C_{4}\times 5C_{1}+9C_{5}}{14C_{5} }\)
\( = \frac{84\times 10+126\times 5+126}{2002}\)

\( = \frac{840+630+126}{2002}\)

\( = \frac{1596}{2002}\)

\( = \frac{114}{143}\)

Total Probability will be,
\(=\frac{(6C1 \times 5C4)+(6C2 \times 5C3)+ (6C3\times 5C2)+(6C4 \times 5C1 }{(11C5)}\)
\(= \frac{455}{462}\).

• The three axioms of axiomatic approach of probability. They are,
• The probability of an event ranges from 0 and 1.
• The sum of probabilities of all sample points of the sample space is equal to 1.
• If A and B are mutually exclusive events, then the probability of occurance of either A or B shall be P (A B) = P(A) + P(B).

Required probability = \( \frac{^{12}C_2 +^{36}C_2 }{ ^{52}C_2} =\frac{116}{221} \)

Sample space of a dice roll = {1, 2, 3, 4, 5, 6}

A = {1, 2, 3, 4}

B = {1, 3, 5}

C = {3, 6}

A ∩ B ∩ C = {3}

Answer: A ∩ B ∩ C = {3}

Let the number of white balls be 'w', then

\(\frac{w}{w+10}=\frac{3}{8}\)

\(w =6\)

Probability of getting 2 red balls

\(=\frac{{ }^{10} C _{2}}{{ }^{16} C _{2}}=\frac{3}{8}\)

Hence, option D is correct.

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

Odd no. of tails = {TTT,HHT,HTH,THH}

Probability = 4/8 = 1/2

There are total 2 red kings and 26 red cards.

Probability of red king \(=\frac{2}{52}\)

Remaining cards \(=51\)

Remaining red cards \(=25\)

Probability of red card \(=\frac{25}{51}\)

Required \(P=\frac{2}{52} \times \frac{25}{51}= \frac{25}{1326}\)

\(\therefore\) The answer is \( \frac{25}{1326}\).

Here total of 21 students are present , at most two boys means there can be no boys , 1 boy and 2 girls and 2 boys and a girl.

But according to probability , this can be also written as \( 1 \) - P ( All three are boys).

So P( All three are boys ) = \( 12C_{3} \div 21C_{3} \)

= \(\frac{22}{133} \)

So required probability is \( 1 - \frac{22}{133}\)

= \(\frac{111}{133} \)

Probability of one basket =1/2

1st Basket Pink toy probability = \(\frac{1}{2}\times \)(\(6 C_1/14 C_1\))

2nd Basket Pink toy probability = \(\frac{1}{2}\times (8 C_1/15 C_1) \)

Adding both \(\frac{1}{2}\times \frac{1}{2}\) + \(\frac{1}{2}\times \frac{4}{5}\)

\( \frac{3}{14}+ \frac{4}{15}\)= \(\frac{101}{210}\)

Since we want exactly 2 of them to be children, we choose the 2 children first in 4 C 2 =6 ways

Choose the other 2 people from the men and women in 5C 2 =10 ways

The universal set had 9C4=126 number of ways

Hence, probability is = \( \frac{10×6}{126}\)= \(\frac{10}{21}\)

The number of success are in a sequence of n independent experiments each asking yes - i.e., whether the outcome is even or odd. Therefore, the given distribution is binomial distribution. As we know that, the variance for the binomial distribution is given by,,

\(\sigma ^{2} =npq\)

Where, \(p\)= The success probability \( q\)= The failure probability, \(n\)= The number of experiments.

Here ,\(n=5\)

\(p \),getting an odd number =\(\frac{3}{6} =\frac{1}{2}\)

\(q =1-p\)

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

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

Variance,\(\sigma ^{2} =5\times\frac{1}{2}\times\frac{1}{2}\)

=\(\frac{5}{4}\)

• Probability of an event + probability of its complementary event = 1
• P(A) + probability of complementary = 1
• Probability of a complement = 1 - P(A)

A number has exactly 3 factors if the number is squares of a prime number.

Squares of 11, 13, 17, 19, 23, 29, 31 are 3-digit numbers.

Hence, the required probability is \(\frac{7}{900}\)

.

\(^8C_3.P^3(1-P)^5\)

=\(\frac1{25}^8C_5P^5(1-P)^3\)

=25(1-p)\(^{2}\)= \(p^{2}\)

=\(\frac{p}{1-p}\)= \( \pm\)5

solving p for +5 and -5

we get p = \(\frac{5}{6}\), \(\frac{5}{4}\) (\(\because p< 1 \))

So the value of p=\(\frac{5}{6}\)

Let 2 events A and B

Odds against A are 1 : 3

So probability of occurrence of A = \(\frac{3}{1+3}\)= \(\frac{3}{4}\)

And non-occurrence of A = \(\frac{1}{4}\)

Odds in favor of B are 2 : 5

So probability of occurrence of B =\(\frac{2}{2+5}\) =\(\frac{2}{7}\)

And non-occurrence of B = \(\frac{5}{7}\)

Case 1: A occurs and B does not occur So probability =\(\frac{3}{4}\) \(\times\)\(\frac{5}{7}\) = \(\frac{15}{28}\)

Case 2: B occurs and A does not occur

So probability = \(\frac{2}{7}\) \(\times\) \(\frac{1}{4}\) = \(\frac{2}{28}\)

So probability that one will occur

=\(\frac{15}{28}\)+ \(\frac{2}{28}\)

= \(\frac{17}{28}\)

The probability of getting odd number is = \(\frac{1}{2}\)

The probability of getting even number is = \(\frac{1}{2}\)

The probability of getting one odd and one even in two dice throw =\(2\times \frac{1}{2}\times\frac{1}{2} \) = \(\frac{1}{2}\)

If the balls are being drawn one after another, the probability of drawing any color of balls for every fresh draw changes.

∴ Required probability = \( \frac{10}{16} \times \frac{9}{15} \times \frac{8}{14} = \frac{5}{8} \times \frac{3}{5} \times \frac{4}{7} = \frac{3}{14} \)

Hence, option A is correct.

Required probability\(=\frac{ ^3C_1\times^5C_1\times^4C_2}{^{12}C_4}=\frac{2}{11}\)

So probability of occurance of event = 3/(3+5) = 3/8

Total number of ways of selecting 3 people from 10 people = \(10 C_3=120\) ways
Required = \(4 C_1 \times 6 C_2 \) ways = 60 ways
Probability = \( \frac{60}{120}= \frac{1}{2}\)

• Let A be the event of selecting a colour blind person and Ei be the event of selecting a person.
• Then P(A) = P(E1) P(A|E1) + P(E2) P(A|E2), =(\( \frac{1}{2}\))(\( \frac{5}{100}\)) + (\( \frac{1}{2}\))(\( \frac{10}{250}\)) = 0.045

• Formula for the probability of A and B (independent events): p(A and B) = p(A) * p(B).
• If the probability of one event doesn't affect the other, you have an independent event.
• All you do is multiply the probability of one by the probability of another.

Probability of getting specific card from first deck = \( \frac{1}{52} \) . Probability of getting specific card from second deck = \( \frac{1}{52} \) .
So required probability = \( \frac{1}{52} \times \frac{1}{52} = \frac{1}{2704} \) .

P(odd) =P (even) = 1/2 (because there are 50 odd and 50 even numbers)

Sum of the three numbers can be odd only under the following 4 scenarios:

Odd + Odd + Odd = 1/21/21/2 =1/8

Odd + Even + Even =1/21/21/2 =1/8

Even + Odd + Even = 1/21/21/2 =1/8

Even + Even + Odd =1/21/21/2 =1/8

Adding all 1/8+1/8+1/8+1/8=4/8= 1/2.

Red \(=0.4 \times 30=12\)

Green \(+\) Blue \(=30-12=18\)

Green: Blue \(=4: 5\)

So, Green balls \(=8\) and Blue balls \(=10\)

Probability of first ball not being red \(=\frac{18}{30}\)

Probability of second ball not being blue \(=\frac{20}{30}\)

Probability \(=\frac{18}{30} \times \frac{20}{30}=\frac{2}{5}\)

Hence, option A is correct.

(1) Favourable cases of a sum 8 are : \((2,6),(3, 5) (4, 4)(5 ,3)(6,2)\)

(2) Favourable cases for a number 3 are : \((3, 5),(5 ,3)\)

\(P(A)\)= Probability of getting 3 at least once

\(P(B)\)= Probability of getting sum to be 8

\(P(A/B)=\) \(\frac{P(A \cap B)}{P(B)}= \frac{ \frac{2}{36}}{\frac{5}{36}}=\) \(\frac{2}{5}\)

Number of boys = 5 and number of girls = 3
Total possible outcomes=\(8_{C_{3}}\)
Favourable outcome(more girle than boys)=\(5_{C_{0}}\times 3_{C_{3}}+5_{C_{1}}\times 3_{C_{2}}\)=16
Required probability=\(\frac{16}{8_{C_{3}}}\)
=\(\frac{16}{8\times 7}\)
=\(\frac{2}{7}\)

Probability of choosing a bag \(=\frac{1}{2}\)
Probability of choosing two red balls from Bag - \(A=\frac{{ }^{7} C_{2}}{{ }^{15} C_{2}}=\frac{21}{105}=\frac{1}{5}\)
Probability of choosing two red balls from Bag - B \(=\frac{x_{c_{2}}}{x+7 c_{2}}=\frac{x(x-1)}{(x+6)(x+7)}\)
ATQ
\(\frac{2}{15}=\frac{1}{2}\left[\frac{1}{5}+\frac{x(x-1)}{(x+6)(x+7)}\right]\) \(\mathrm{x}=3,-1\)
So, required numbers of balls is 3 as numbers of balls cannot be negative.

Let the number of blue balls \(=x\)
And the number of red balls \(=x+4\)
Number of yellow balls \(=x-1\)
\(x+x+4+x-1=15\)
\(3x=12\)
\(x=4\)
Required probability \(=\frac{^8C_3}{^{15}C_3}\)
\(=\frac{8}{65}\)

The only even prime number = 2

Since 2 is not present probability = 0

The possible outcomes of the random experiment is

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

Let \(E\) denote the event getting an odd number

Then \(E=\{1,3,5\}\)

\(F\) denote the event getting a number greater than \(3\). Then

\(F=\{4,5,6\}\)

Now \(P(E)=\frac{3}{6}=\frac{1}{2} \ \ , \ \ P(F)=\frac{3}{6}=\frac{1}{2}\)

\(E \cap F=\{5\}\)

\(P(E \cap F)=\frac{1}{6}\)


\(\hspace{20mm}=1/2+1/2-1/6\)

\(\hspace{20mm}=1-1/6\)

\(\hspace{20mm}=5/6\)

According to Alden the first dice shows an even number so P(First ) = \( \frac{1}{2}\)
and second dice shows a number less that 5 , so P(Second) = \( \frac{2}{3} \) .
Third dice shows an odd number so P(Third ) = \( \frac{1}{2} \) . So required probability will be \( \frac{1}{2} \times \frac{1}{2} \times \frac{2}{3} = \frac{1}{6} \) .

Let P denote perfect toy and D defective toy.

Sample space = { PPP, PPD, PDP, DPP, DDD, DDP, DPD, PDD }

n(Sample Space) = 8

Atleast 2 perfect toys = { PPP, PPD, PDP, DPP }

n(Atleast 2 Perfect toys) = 4

P(Box passes) = P(Atleast 2 Perfect toys)

= \(\frac{4}{8}\)

⇒ 0.5

∴ Probability of box passing is 0.5

We have, n (S) = 52

The event of getting a black king = 2

So, n (E) = 2

P(E) = n(E) / n(S) = 2 / 52 = 1 / 26

n(S) =36

E= event of black of die

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

P(E)=\(\frac{6}{36}\)=\(\frac{1}{6}\)

F= getting a sum greater than 9

F = {(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)}

E \(\cap\)F={(5,5),(5,6)}

P(E \(\cap\)F) = \(\frac{2}{36}\)=\(\frac{1}{18}\)

required probability P(E/F) = \(\frac{P(E\cap F)}{P(F)}\)=\(\frac{\frac{1}{18}}{\frac{1}{6}}\)=\(\frac{1}{3}\)

• Number of trails, n= 20

  number of defective toycars, r= 2

• Probability of getting defective toycars in a single trial= \(\frac{50}{100}= \frac{1}{2}\)

• Then the probability of getting good toycars in a single trial= \(\frac{50}{100}= \frac{1}{2} \)

• The probability of defective P(r)= \(_{r}^{n}\textrm{C} \) \(p^{r}\)\(q^{n- r}\)

  =\(_{2}^{20}\textrm{C} \)\((\frac{1}{2})^{2}\)\((\frac{1}{2})^{20-2}\)

  =\(\frac{19\times 10\times1}{4\times 64\times 64}\)

  = 0.0116

\( Number\: of \:possible\: combination\: of\: 4\: persons\: in\: which\: 3\: have\: to\: be\: men\:is,\\\\\ (3 \:men\: out\: of \:7 \:\times 1 \:Women\: out \:of \:6)\: or\: (4\: men\: out\: of\: 7)\\\\ =\:7C_{3}\times 6C_{1}+7C_{4}\\\\ Total \:possible\: outcomes =\:13C_{4}\\\\ Required\:probability=\:\frac{7C_{3}\times 6C_{1}+7C_{4}}{13C_{4}}\\\\ =\:\frac{\frac{7!}{3!\times 4!}\times 6+\frac{7!}{4!\times 3!}}{\frac{13!}{4!\times 9!}}\\\\ =\:\frac{35\times 6+35}{715}\\\\ =\:\frac{245}{715}\\\\ =\:\frac{49}{143} \)

Here S = {HH, HT, TH, TT}.
Let E = event of getting at least one head = {HH, HT, TH}.
Therefore P(E) = \( \frac{N(E))}{n(S))} \) = \( \frac{3}{4} \)

Two cases are possible

1) \(\frac{^4_4C \times ^5_1C}{^9_5C}\)

2)\(\frac{^4_3C \times ^5_2C}{^9_5C}\)

Required Probability = \(\frac{^4_4C \times ^5_1C}{^9_5C}\)+\(\frac{^4_3C \times ^5_2C}{^9_5C}\) = \(\frac{5+40}{120} = \frac{5}{14}\)

So, option (d) is the correct answer.

Required probability\(=\frac{^{20}C_2\times^{10}C_2}{^{30}C_4}=\frac{190}{609}\)

Probability of getting \(1^{\text {st }}\) red ball \(=\frac{x}{(x+4+5)}=\frac{x}{(x+9)}\)

Probability of getting \(2^{\text {nd }}\) red ball without replacement \(=\frac{(x-1)}{(x+8)}\)

Probability of getting \(3^{\text {rd }}\) red ball without replacement \(=\frac{(x-2)}{(x+7)}\)

According to the question,

\(\left[\frac{x}{(x+9)}\right] \times\left[\frac{(x-1)}{(x+8)}\right] \times\left[\frac{(x-2)}{(x+7)}\right]=\frac{1}{16}\)

Now we will check through option and hence, option \(A =7\) satisfied the equation

So, value of \(x=7\).

• In the probability function of the Binomial distribution, ‘x’ stands for the number of successes.
• ’p’ represents probability for success in a single trial and q represents probability for failure.
• \(q = 1-p\).
• Hence \(p+q =1 \).

After taking away respective balls, Number of balls in the box = \(75+25+50=150\)
Percentage of black balls = \(\frac{50}{150}\times 100= \frac{100}{3} = 33\frac{1}{3}\) %

Total cards = 52.

Number of ways of drawing 2 cards = \(52C_{2}\).

Honours card – Ace, King, Queen and jack.

Total honor cards = 16

Number of ways of drawing 2 honor cards = \(16C_{2}\).

Therefore, Required probability = \(\frac{16C_{2}}{52C_{2}}\)

= \(\frac{16\times 15}{52\times 51}\)

= \(\frac{4\times 5}{13\times 17}\)

= \(\frac{20}{221}\)

Hence, option C is correct.

Let the probability of getting an even number = 'x'

Then, probability of getting an odd number \(=\left\{x+\left(\frac{1}{9}\right)\right\}\)

So, \(x+\left\{x+\left(\frac{1}{9}\right)\right\}=2 x+\left(\frac{1}{9}\right)=1\)

Or, \(x=\left\{1-\left(\frac{1}{9}\right)\right\} \div 2=\left(\frac{4}{9}\right)\)

So, probability of getting an even number and an odd number on the die are \(\frac{4}{9}\) and \(\frac{5}{9}\), respectively

Since there are 3 odd numbers and 3 even numbers on the die,

Probability of getting a specific even number \(=\frac{4}{9} \times \frac{1}{3}=\frac{4}{27}\)

Similarly, probability of getting a specific odd number \(=\frac{5}{27}\)

On throwing two dice at once, all possible events where the sum is greater than or equal to 9 are \(=\{(3,6),(4,5)(4,6),(5,4),(5,5),(5,6),(6,3)(6,4),(6,5)\) and \((6,6)\}\)

So, required probability \(=3 \times \frac{4}{27} \times \frac{4}{27}+6 \times \frac{5}{27} \times \frac{4}{27}+\frac{5}{27} \times\) \(\frac{5}{27}=\frac{193}{729}\)

Let A = three black balls are drawn
\(E_i\) = Bag contains i white and 10 – i black balls \(P(E_1/A)=\frac{P(A|E_1)P(E_1)}{\sum_{i=0}^{10}P(A|E_i)P(E_i)}\) \(=\frac{\frac{9C_3}{11\times 10C3}}{\frac{1}{11}(\frac{10C3+9C3+8C3+.......+3C3}{10C3})}\) \(=\frac{9C3}{11C4}\) \(=\frac{14}{55}\)