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svp [43]
3 years ago
9

a red ball and 4 white ball are in a box if two balls are drawn without replacement ,what is the probability of getting a red ba

ll on the first drawn and a white ball on the second​
Mathematics
1 answer:
BARSIC [14]3 years ago
3 0

The probability is 1/5 to get a red ball in 1st draw and a white ball in 2nd draw.

<u>Step-by-step explanation:</u>

  • There are 1 red ball and 4 white balls in a box.
  • The total number of balls in the box = 1 red + 4 white = 5 balls.

The two balls are drawn without replacement.

<u>Drawing the first ball :</u>

The first draw should be a red ball.

The probability to get a red ball = No.of red balls / Total balls in the box.

We know that, No. of red balls is 1 and total balls in the box is 5.

P(red ball) = 1/5

<u>Drawing the second ball :</u>

The second draw should be a white ball.

The probability to get white ball = No.of white balls / Total balls in the box.

We know that,

No. of white balls is 4.

The total balls in the box after the first draw will be 4 balls.

P(white ball) = 4/4

The probability of getting a red ball on the first drawn and a white ball on the second​ draw = P(red ball) × P(white ball)

⇒ (1/5) × (4/4)

⇒ 4/20

⇒ 1/5

Therefore, the probability is 1/5 to get a red ball in 1st draw and a white ball in 2nd draw.

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Identify the coefficients and constants in the expression. 2x – y + 5x
garri49 [273]
I believe the answer is 7x-y
7 0
3 years ago
Instructions: Match each equations with correct type of event.
Mazyrski [523]

The question is incomplete. Below you will find the missing contents.

The correct match of events with order are,

  • P(A)P(B|A) - Dependent event
  • P(A)+P(B) - Mutually exclusive events
  • P(A and B)/P(A) - Conditional events
  • P(A) . P(B) - Independent Events
  • P(A)+P(B) -P(A and B) - not Mutually exclusive events.

When two events A and B are independent then,

P(A and B)=P(A).P(B)

when A and B are dependent events then,

P(A and B) = P(A) . P(B|A)

When two events A and B are mutually exclusive events then,

P(A and B)=0

So, P(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B)

P(A) + P(B) = P(A or B)

When events are not mutually exclusive then the general relation is,

P(A or B) = P(A) + P(B) - P(A and B)

If the probability of the event B conditioned by A is given by,

\mathrm{P(B|A)=\frac{P(A~and~B)}{P(A)}}

Hence the correct match are -

  • \mathrm{P(A)P(B|A)\rightarrow} Dependent event
  • \mathrm{P(A)+P(B)}\rightarrow Mutually exclusive events
  • \mathrm{\frac{P(A ~and ~B)}{P(A)}\rightarrow} Conditional events
  • \mathrm{P(A) . P(B)\rightarrow} Independent Events
  • \mathrm{P(A)+P(B) -P(A ~and ~B)\rightarrow} not Mutually exclusive events.

Learn more about Probability of Events here -

brainly.com/question/79654680

#SPJ10

5 0
2 years ago
Someone plz help I rlly need to finish
bekas [8.4K]

Since TSQ and QSR are supplementary, they give 180 when summed. TSQ is 150, so QSR must be 30.

Therefore, you have

\tan(30)=\dfrac{\sin(30)}{\cos(30)}=\dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \dfrac{1}{2}\cdot\dfrac{2}{\sqrt{3}}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}

6 0
3 years ago
A small theater sold 72 tickets for a
finlep [7]

Answer:

Adult tickets sold= 9x4 = 36

Senior tickets sold = 9x3= 27

Child tickets sold = 9 x 1 = 9

Step-by-step explanation:

If the parts for adult tickets changed within the 2 ratios given, we should have equated Ratio further. But here that is not 5e case. So, put both ratios together:

Senior:Adult:Child

3:4:1

Now,

Total number of parts 3+4+1=8

Number of tickets sold per part = 72/8 = 9

Therefore number of

Adult tickets sold= 9x4 = 36

Senior tickets sold = 9x3= 27

Child tickets sold = 9 x 1 = 9

Hope this helps.

Good Luck

5 0
3 years ago
1. If BCDE is congruent to OPQR, then BC is congruent to<br> A. OR<br> B. OP<br> C. PQ<br> D. QR
Zepler [3.9K]
D just got done with test
8 0
2 years ago
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