Probability - Class 11 Commerce Applied Mathematics - Extra Questions

We have,

number of cases of getting a yellow ball when drawn is $$7$$

$$\begin{array}{l} P\left( { { E_{ 1 } } } \right) =\dfrac { { 2000 } }{ { 5000 } } =0.4 \\ P\left( { { E_{ 2 } } } \right) =\dfrac { { 3000 } }{ { 5000 } } =0.6 \\ P\left( { \dfrac { A }{ { { E_{ 1 } } } }  } \right) =0.01 \\ P\left( { \dfrac { A }{ { { E_{ 2 } } } }  } \right) =0.02 \\ P\left( { \dfrac { { { E_{ 2 } } } }{ A }  } \right) =\dfrac { { P\left( { { E_{ 1 } } } \right) P\left( { \dfrac { A }{ { { E_{ 2 } } } }  } \right)  } }{ \begin{array}{l} P\left( { { E_{ 1 } } } \right) P\left( { \dfrac { A }{ { { E_{ 1 } } } }  } \right) +P\left( { { E_{ 2 } } } \right) P\left( { \dfrac { A }{ { { E_{ 2 } } } }  } \right)  \\ =\dfrac { { 0.6\times 0.02 } }{ { 0.6\times 0.02+0.4\times 0.01 } }  \\ =\dfrac { { 0.012 } }{ { 0.012+0.004 } }  \\ =\dfrac { 3 }{ 4 }  \\  \end{array} }  \\  \end{array}$$

Sum of 2 Numbers

X = { 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 ,12 )

$$\displaystyle p(x) = \left\{ {{1 \over {36}},{2 \over {36}},{3 \over {36}},{4 \over {36}},{5 \over {36}}} \right\}$$

$$E(x) = \sum\limits_{}^{} {xi(p(x) = x)} $$

$$\displaystyle = 2 \times {1 \over {36}} + 3 \times {2 \over {36}} + 4 \times {3 \over {26}} + 5 \times {4 \over {36}} + 6 \times {5 \over {36}} + 7 \times {6 \over {36}} + 8 \times {5 \over {36}} + 9 \times {4 \over {36}} + 10 \times {3 \over {36}} + 11 \times {2 \over {36}} + 12 \times {1 \over {36}}$$

$$\displaystyle{{2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12} \over {36}}$$

$$\displaystyle{{252} \over {36}} = 7$$

Total number of boxes $$=10$$

Total number of balls $$=x+y+z$$

An experiment in which all possible outcomes are known, and the exact outcome cannot be predicted in advance, is called a random experiment.

An event which is impossible to occur, is called an impossible event. The probability of impossible event is always zero.

The total no.of cars $$29+26+23+17+5=100$$

It is impossible.

We have $$\displaystyle p\left( A \right) =p\left( A|B \right) =\frac { p\left( A\cap B \right)  }{ p\left( B \right)  } \Rightarrow p\left( A\cap B \right) =p\left( A \right) p\left( B \right) $$

Let $$E$$ be the event 'Red on first draw', $$F$$ be the event 'Red on second draw' and $$G$$ be the event 'white on second draw',
$$\displaystyle P\left ( E \right )=\frac{6}{10}, P\left ( F \right )=\frac{6}{10}, P\left ( G \right )=\frac{4}{10}$$
$$\displaystyle P\left ( E\cap F \right )=\frac{^{6}P_{2}}{^{10}P_{2}}=\frac{1}{3}$$
$$\displaystyle P\left ( E \right ).P\left ( F \right )=\frac{3}{5}\times \frac{3}{5}=\frac{9}{25}\neq \frac{1}{3}$$
$$\therefore $$ $$E$$ and $$F$$ are not independent

Let $$A=$$the event that the rose bush has withered
Let $$A_{1}=$$the event that the gardener did not water.
    $$A_{2}=$$the event that the gardener watered.
By Bayes's theorem required probability,
$$\displaystyle P\left ( A_{1}/A \right )=\frac{P\left ( A_{1} \right ).P\left ( A/A_{1} \right )}{P\left ( A_{1} \right ).P\left ( A/A_{1} \right )+P\left ( A_{2} \right ).P\left ( A/A_{2} \right )}$$          (i)
Given, $$\displaystyle P\left ( A_{1} \right )=\frac{2}{3}$$
$$\therefore $$ $$\displaystyle P\left ( A_{2} \right )=\frac{1}{3}$$
$$\displaystyle P\left ( A/A_{1} \right )=\frac{3}{4}, P\left ( A/A_{2} \right )=\frac{1}{2}$$
From (i), $$\displaystyle P\left ( A_{1}/A \right )=\frac{\frac{2}{3}.\frac{3}{4}}{\frac{2}{3}.\frac{3}{4}+\frac{1}{3}.\frac{1}{2}}=\frac{6}{6+2}=\frac{3}{4}$$

Let $$E$$ be the event 'Red on first draw', $$F$$ be the event 'Red on second draw' and $$G$$ be the event 'white on second draw',
$$\displaystyle P\left ( E \right )=\frac{6}{10}, P\left ( F \right )=\frac{6}{10}, P\left ( G \right )=\frac{4}{10}$$
$$\displaystyle P\left ( E \right ).P\left ( G \right )=\frac{6}{10}\times \frac{4}{10}=\frac{6}{25}$$
$$\displaystyle P\left ( E\cap G \right )=\frac{^{6}P_{1}\times ^{4}P_{1}}{^{10}P_{2}}=\frac{4}{15}$$
$$\therefore $$ $$P\left ( E \right ).P\left ( G \right )\neq P\left ( E\cap G \right )$$
$$\therefore $$ $$E$$ and $$G$$ are not independent

$$\displaystyle P\left ( A \right )=P\left ( B \right )=P\left ( C \right )=\frac{2}{4}=\frac{1}{2}$$
Also $$\displaystyle P\left ( A\cap B \right )=\frac{1}{4}=P\left ( A \right )P\left ( B \right ), P\left ( A\cap C \right )=\frac{1}{4}=P\left ( A \right )P\left ( C \right ), P\left ( B\cap C \right )=\frac{1}{4}=P\left ( B \right )P\left ( C \right )$$
but $$P\left ( A\cap B\cap C \right )=0\neq P\left ( A \right )P\left ( B \right )P\left ( C \right )$$
$$\therefore $$ $$A, B$$ & $$C$$ are not independent

Let $$A$$ be the event of drawing a red card when one card is drawn out of $$51$$ cards (excluding missing card.) Let $$A_{1}$$ be the event that the missing card is red and $$A_{2}$$ be the event that the missing card is black.
Now by Bayes's theorem, required probability,
$$\displaystyle P\left ( A_{1}/A \right )=\frac{P\left ( A_{1} \right ).\left ( P\left ( A/A_{1} \right ) \right )}{P\left ( A_{1} \right ).P\left ( A/A_{1} \right )+P\left ( A_{2} \right ).P\left ( A/A_{2} \right )}$$          (i)
In a pack of $$52$$ cards $$26$$ are red and $$26$$ are black.
Now $$P\left ( A_{1} \right )=$$probability that the missing card is red$$\displaystyle =\frac{^{26}\textrm{C}_{1}}{^{52}\textrm{C}_{1}}=\frac{26}{52}=\frac{1}{2}$$
$$P\left ( A_{2} \right )=$$probability that the missing card is black$$\displaystyle =\frac{26}{52}=\frac{1}{2}$$
$$P\left ( A/A_{1} \right )=$$probability of drawing a red card when the missing card is red.
   $$\displaystyle =\frac{25}{51}$$
[$$\because $$ Total number of cards left is $$51$$ out of which $$25$$ are red and $$26$$ are black as them is sing card is red]
Again $$P\left ( A/A_{2} \right )=$$Probability of drawing a red card when the missing card is back$$\displaystyle =\dfrac{26}{51}$$
Now from (i), required probability, $$\displaystyle P\left ( A_{1}/A \right )=\dfrac{\dfrac{1}{2}.\dfrac{25}{51}}{\dfrac{1}{2}.\dfrac{25}{51}+\dfrac{1}{2}.\dfrac{26}{51}}=\dfrac{25}{51}$$

We can think of the following possibilies
(i)    $$E_{1}$$    6W     0 B
(ii)   $$E_{2}$$      6W     1 B
(iii)  $$E_{3}$$     6W     2 B
(iv)  $$E_{4}$$     6W     3 B
Clearly $$\displaystyle P\left ( E_{1} \right )=P\left ( E_{2} \right )=P\left ( E_{3} \right )=P\left ( E_{4} \right )=\dfrac {1}{4}$$
Let $$'A'$$ be the event that two balls drawn one by one with replacement are both white therefore we
have to find $$\displaystyle P\left ( \dfrac {E_{4}}{A} \right )$$
By Baye's theorem $$\displaystyle P\left ( \dfrac {E_{4}}{A} \right )=\dfrac {P\left ( \dfrac {A}{E_{4}} \right )\times P\left ( E_{4} \right )}{P\left ( \dfrac {A}{E_{1}} \right )\times P\left ( E_{1} \right )+P\left ( \dfrac {A}{E_{2}} \right ).P\left ( E_{2} \right )+P\left ( \dfrac {A}{E_{3}} \right ).P\left ( E_{3} \right )}+P\left ( \dfrac {A}{E_{4}} \right ).P\left ( E_{4} \right )$$
$$\displaystyle P\left ( \dfrac {A}{E_{4}} \right )=\dfrac {6}{9}\times \dfrac {6}{9}; P\left ( \dfrac {A}{E_{3}} \right )=\dfrac {6}{8}\times \dfrac {6}{8}; P\left ( \dfrac {A}{E_{2}} \right )=\dfrac {6}{7}\times \dfrac {6}{7}; P\left ( \dfrac {A}{E_{1}} \right )=\dfrac {6}{6}\times \dfrac {6}{6};$$
Putting values $$\displaystyle P\left ( \dfrac {E_{4}}{A} \right )=\dfrac {\dfrac {1}{81}}{\dfrac {1}{81}+\dfrac {1}{64}+\dfrac {1}{49}+\dfrac {1}{36}}=0.162$$

Let $$E$$ and $$F$$ be the events 'Class $$XI$$ is chosen' and 'Class $$XII$$ is chosen' respectively,
Then $$\displaystyle P\left ( E \right )=\frac{2}{5}, P\left ( F \right )=\frac{3}{5}$$
Let $$A$$ be the event 'Student chosen is brilliant'.
Then $$\displaystyle P\left ( A/E \right )=\frac{5}{50}$$ and $$\displaystyle P\left ( A/F \right )=\frac{8}{50}$$
$$\therefore $$ $$\displaystyle P\left ( A \right )=P\left ( E \right ).P\left ( A/E \right )+P\left ( F \right ).P\left ( A/F \right )=\frac{2}{5}.\frac{5}{50}+\frac{3}{5}.\frac{8}{50}=\frac{34}{250}.$$
$$\therefore $$ $$\displaystyle P\left ( E/A \right )=\frac{P\left ( E \right ).P\left ( A/E \right )}{P\left ( E \right )P\left ( A/E \right )+P\left ( F \right ).P\left ( A/F \right )}=\frac{5}{17}.$$

When dice is thrown, number of observations in the sample space $$= 6 \times 6 = 36$$
Let $$A$$ be the event that the sum of the numbers on the dice is $$4$$ and $$B$$ be the event that the two numbers appearing on throwing the two dice are different.
$$\therefore A=\left\{ \left( 1,3 \right) ,\left( 2,2 \right) ,\left( 3,1 \right)  \right\} $$
$$B=\left\{ \left( 1,2 \right) ,\left( 1,3 \right) ,\left( 1,4 \right) ,\left( 1,5 \right) ,\left( 1,6 \right) \\ \left( 2,1 \right) ,\left( 2,2 \right) ,\left( 2,3 \right) ,\left( 2,4 \right) ,\left( 2,5 \right) ,\left( 2,6 \right) \\ \left( 3,1 \right) ,\left( 3,2 \right) ,\left( 3,3 \right) ,\left( 3,4 \right) ,\left( 3,5 \right) ,\left( 3,6 \right) \\ \left( 4,1 \right) ,\left( 4,2 \right) ,\left( 4,3 \right) ,\left( 4,4 \right) ,\left( 4,5 \right) ,\left( 4,6 \right) \\ \left( 5,1 \right) ,\left( 5,2 \right) ,\left( 5,3 \right) ,\left( 5,4 \right) ,\left( 5,5 \right) ,\left( 5,6 \right) \\ \left( 6,1 \right) ,\left( 6,2 \right) ,\left( 6,3 \right) ,\left( 6,4 \right) ,\left( 6,5 \right) ,\left( 6,6 \right)  \right\} $$
$$A\cap B=\left\{ \left( 1,3 \right) ,\left( 3,1 \right)  \right\} $$
$$\therefore P\left( B \right) = \displaystyle\frac { 30 }{ 36 } = \displaystyle\frac { 5 }{ 6 } $$ and $$P\left( A\cap B \right) = \displaystyle\frac { 2 }{ 36 } = \displaystyle\frac { 1 }{ 18 } $$
Let $$P\left( A|B \right) $$ represent the probability that the sum of the numbers on the dice is $$4$$, given that the two numbers appearing on throwing the two dice are different.
$$\therefore P\left( A|B \right) = \displaystyle\frac { P\left( A\cap B \right)  }{ P\left( B \right)  } = \displaystyle\frac { \displaystyle\frac { 1 }{ 18 }  }{ \displaystyle\frac { 5 }{ 6 }  } = \displaystyle\frac { 1 }{ 15 } $$
Therefore, the required probability is $$\displaystyle\frac { 1 }{ 15 } $$

P(g) = probability of guessing $$= \dfrac{1}{3}$$

P(c) = probability of copying $$= \dfrac{1}{6}$$

P(k) = probability of knowing $$= 1 - \dfrac{1}{3} - \dfrac{1}{6}= \dfrac{1}{2}$$

Probability of answering correctly during guess $$= 1/4$$ (one out of four choices)

Probability of answering correctly during copy $$= 1/8$$ 

Probability of answering correctly when he knew answer to the question (by solving) = 1 (As he knows answer he will answer it correctly).

Hence using Bayes formula (or theorem)

The probability that he knew the answer to the question, given that he answered it correctly

$$= \dfrac{\left(1 \times  \dfrac{1}{2}\right)}{\left(\dfrac{1}{3} \times  \dfrac{1}{4} + \dfrac{1}{6} \times  \dfrac{1}{8} + \dfrac{1}{2}\times  1\right)} $$

$$= \dfrac{\dfrac{1}{2}}{\left(\dfrac{1}{12}+ \dfrac{1}{48} + \dfrac{1}{2} \right) }= \dfrac{24}{29}$$

(a) Out of six possibilities only once the number $$4$$ occurs hence probability is $$\dfrac {1}{6}$$. Similarly we fill the remaining

Outcome	1	2	3	4	5	6
Probability (P)				1/6

(b) The sum of all probabilities
$$P\left(1\right) + P\left(2\right) + P\left(3\right) + P\left(4\right) + P\left(5\right) + P\left(6\right)$$
$$=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=1$$

When we toss a coin, the possible outcomes are only two - head or tail.

Let $${E_1},{E_2},{E_3}$$  be the event of choosing a card red on both sides , card blue on both sides and blue on side & red on other side 

$$\textbf{Find the required number of ways}$$

Let us consider the probabilities of production 

Let the poor student $$P\left( A \right)=35%$$

Meritorious student $$P\left( B \right)=20%$$

Both  $$P\left( A\cap B \right)=15%$$

Now,

$$P\left( A \right)=\dfrac{35}{100}$$

$$P\left( B \right)=\dfrac{20}{100}$$

$$P\left( A\cap B \right)=\dfrac{15}{100}$$

(a).He is poor if it is know that he meritorious

Probability

$$ P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}=\dfrac{\dfrac{15}{100}}{\dfrac{20}{100}} $$

$$ P\left( \dfrac{A}{B} \right)=\dfrac{15}{20} $$

$$ P\left( \dfrac{A}{B} \right)=\dfrac{3}{4} $$

 

(a).He is meritorious if it is know that he poor

Probability

$$ P\left( \dfrac{B}{A} \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}=\dfrac{\dfrac{15}{100}}{\dfrac{35}{100}} $$

$$ P\left( \dfrac{A}{B} \right)=\dfrac{15}{35} $$

$$ P\left( \dfrac{A}{B} \right)=\dfrac{5}{7} $$

Hence, this is the answer.

This is a conditional probability so we use Bayes Law.
P(A | B) = P(B | A) $$\times$$ P(A)/ P(B).

Where A = all balls white
B = draw 2 white balls

P(B) is not given. I will assume there are 3 possible worlds: 4 whites, 3 white and 1 non white,
and 2 whites and 2 non-white. I need to also know the probabilities for each of these worlds. I will assume they are equal, the standard assumption if no information is given.

P(B) = P(B| 4W)$$\times$$P(4W) +P(B| 3W, 1non-white)*P(3W, 1 non-white) + P(B | 2white,2 non-white).

P(B|4W) = 1
P(4W) = 1/3 = P(3W and 2 non) = P(2W and 2 non)

P(B | 3W and 1 non) = 3/4$$\times$$2/3 = 1/2

P(B | 2W and 1 non) = 2/4$$\times$$1/3 = 1/6

Putting this all together with Bayes:
P(4W in bag | drew 2 white) = 1$$\times$$ 1/3/ ( 1$$\times$$1/3 + 1/2$$\times$$1/3 + 1/6$$\times$$1/3) = 3/5

Let

Consider the problem

Let $${E}_{1},{E}_{2}$$ and$${E}_{3}$$ be the events of the insured vehicle i.e., scooter, car and truck

Let $${E_1}\,,{E_2}$$ be the event of choosing bag 1 and bag 2 respectively.

If the events are mutually exclusive 

The outcomes are not equally likely since the area covered by regions is not same .

Foraloadeddie,itisgiventhat

$$p(1)=p(2)=0.2$$

$$p(3)=p(6)=0.1$$ and $$p(4)=0.3$$

Also,dieisthrowntwotimesHere,A=SamenumbereachtimeandB=Totalscoreis10ormore

$$\therefore A=(1,1),(2,2)(3,3)(4,4)(5,5)(6,6)$$

sample space $$[MFS,MSF,SMF,SFM,FMS,FSM]$$
$$P(E)=\dfrac{4}{6}$$
$$P(F)=\dfrac{2}{6}$$
$$P(E \cap F)=\dfrac{2}{6}$$
$$P(E|F)=\dfrac{P(E \cap F)}{P(F)}=\dfrac{\dfrac{2}{6}}{\dfrac{2}{6}}=1$$

It is possible but not certain.

It can happen but not certain.

Let, 

The total number of elementary events associated to the random experiment of throwing these dice simultaneously is $$6\times 6\times 6=216$$

When a coin is tossed, there are two possible outcomes, head$$(H)$$ and tail $$(T)$$. Hence the sample space in this experiment is,

Given $$A$$ die is thrown twice

When a dice in thrown we get the following outcomes

Out of $$4$$ men and $$6$$ women one person can be chosen in $$^{10}{{C}_{1}}=10$$ ways.

Let A and B denotes the events that a randomly chosen student passes first and second examinations respectively. 

Consider the following events:$$A =$$Integer chosen is a multiple of $$2$$, $$B =$$Integer chosen is a multiple of $$9$$.

Let $$ D_1 $$ denote the event that the patient has disease $$ d_1 $$ . The events $$ D_2 $$ and $$ D_3 $$ are defined similarly . 

Then $$ P(D_1) = \dfrac{3200}{10000} = 0.32 ,\, P(D_2) = \dfrac{3500}{10000} = 0.35 $$ and $$ P(D_3) = \dfrac{3300}{10000} = 0.33 $$ 

 Let $$ S $$ be the event that the patient shows the symptoms $$ S $$ 

Then , $$ P( S / D_1) = \dfrac{P(S \cap D)}{P(D_1)} = \dfrac{3100}{3200} = 0.97 $$ (approx.)

$$ P(S / D_2) = \dfrac{3300}{3500} = 0.94 $$ (approx) and $$ P ( S / D_3) = \dfrac{3000}{3300} = 0.97 $$ ( approx.) 

Using Bayes' theorem , we get 

$$ P(D_1 / S) $$ = The probability that the patient has disease $$ d_1 $$ knowing that he / she ha symptoms $$ S_1 , S_2 ,$$ ...., $$ S_6 $$ 

$$ P(D_1 / S) = \dfrac{P(D_1) P(S / D_1)}{P(D_1)P(S / D_1) + P(D_2) P(S / D_2) + P(D_3) P(S / D_3)} $$ 

                 $$ = \dfrac{0.32 \times 0.97}{0.32 \times 0.97 + 0.35 \times 0.94 + 0.33 \times 0.91} $$ 

                $$ = \dfrac{0.3104}{0.3104 + 0.329 + 0.3003} = \dfrac{0.3104}{0.937} = 0.33 $$ approx. 

 Similarly , $$ P ( D_2 / S) = \dfrac{0.329}{0.9397} = 0.35 $$ approx. 

 $$ P(D_3 / S) = \dfrac{0.3003}{0.9397} = 32 $$ approx. 

Thus , knowing that the patient has symptoms $$ S_1 , S_2, $$ ........., $$ S_6 $$ , the probability that he has disease $$ d_1 $$ is $$ 0.33 $$ , the probability that he has disease $$ d_2 $$ is $$ 0.35 $$ , the probability that he has disease $$ d_3 $$ is $$ 0.32 $$ . Therefore , the doctor should conclude that the patient is most likely to have disease $$ d_2 $$.

Yes

⇒ When a baby child is born. Born baby may be boy or girl. Probability for baby is boy is 50% while baby is girl also 50%. Therefore, it is also an equally likely event.

In throwing a dice , sample space
S = { 1 ,2 , 3 , 4 , 5 ,6}
Then according to question 
E = {4}
and F  = { 2 ,4  6,}
Thus E∩F=4≠ϕE∩F=4≠ϕ
Thus E and F are not mutually exclusive events.

Yes

⇒ In the trial there are only two cases – the first answer may be right or it may be wrong because it is true-false question.

⇒ 50% probability answer is right or 50% probability question is wrong. Therefore, it is an equally likely event.

Let define the following events:
$${ E }_{ 1 }:$$ Examine guesses the answer to the question.
$${ E }_{ 2 }:$$ Examine copies the answer to the question.
$${ E }_{ 3 }:$$ Examine know the answer to the question.
$$E:$$ answer is correct
Here $$\displaystyle P\left( { E }_{ 1 } \right) =\frac { 1 }{ 3 } ,P\left( { E }_{ 2 } \right) =\frac { 1 }{ 6 } $$
$$\displaystyle \therefore P\left( { E }_{ 3 } \right) =1-\left( \frac { 1 }{ 3 } +\frac { 1 }{ 6 }  \right) =\frac { 1 }{ 2 } $$
Since the question is a multiple choice question with four choice so the probability that the answer is correct 
when is guessed is $$\displaystyle P\left( \frac { E }{ { E }_{ 1 } }  \right) =\frac { 1 }{ 4 } $$
Also the probability that his answer is correct, given that he copied it is $$\displaystyle \frac { 1 }{ 8 } $$
i.e $$\displaystyle P\left( \frac { E }{ { E }_{ 2 } }  \right) =\frac { 1 }{ 8 } $$
Moreover his answer is correct, given he knew the answer is a sure event, so $$\displaystyle P\left( \frac { E }{ { E }_{ 3 } }  \right) =1$$
Hence by Baye's theorem the required probability is
$$\displaystyle P\left( \dfrac { { E }_{ 3 } }{ E }  \right) =\dfrac { P\left( \dfrac { E }{ { E }_{ 3 } }  \right) .P\left( { E }_{ 3 } \right)  }{ \sum _{ i=1 }^{ 3 }{ P\left( \dfrac { E }{ { E }_{ i } }  \right) .P\left( { E }_{ i } \right)  }  } =\dfrac { 1.\dfrac { 1 }{ 2 }  }{ \dfrac { 1 }{ 4 } .\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 8 } .\dfrac { 1 }{ 6 } +\dfrac { 1 }{ 2 } .1 } =\dfrac { 24 }{ 29 } $$

When a balance die is thrown device, total no. of out come $$=6\times 6= \boxed {36}$$ 

There are $${ 2 }^{ n }$$ teams total

First bag- 4 white ,5 black