Probability

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31.

There are 4 boys and 4 girls. They sit in a row randomly. What is the chance that all the girls do not sit together?

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Shortcut:
There are 'p' boys and 'q' girls. If they sit in a row randomly, then the chance that
(i) all the girls sit together is given by

(p + 1)!q! / (p + q)!

(ii) all the boys sit together is given by

(q + 1)!p! / (p + q)!

(iii) all the girls do not sit together is given by 1 −

(p + 1)!q! / (p + q)!

(iv) all the boys do not sit together is given by 1 −

(q + 1)!p! / (p + q)!

(v) no two girls sit together (p > q) is given by 1 −

^{(p + 1)}C_q p!q! / (p + q)

Here, p = 4 , q = 4
Using these values in the shortcut (iii), we get:
P(all the girls do not sit together) = 1 −

(4 + 1)!4! / (4 + 4)!

1 −

5!4! / 8!

= 1 −

5! x 4 x 3 x 2 / 8 x 7 x 6 x 5!

= 1 −

1 / 7 x 2

= 1 −

1 / 14

14 − 1 / 14

13 / 14

32.

A box contains 4 black balls, 3 red balls and 5 green balls. 2 balls are drawn from the bos at random. What is the probability that both the balls are of the same colour?

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Shortcut:
A box contains p black balls, q red balls and r green balls. 'z' balls are drawn from the bos at random.
The probability that all the balls are of the same colour is given by

^pC_z + ^qC_z + ^rC_z / ^{(p + q + r)}C_z

, where z < p, q, r
Here, p = 4 , q = 3, r = 5, z = 2
Using these values in the shortcut, we get:

⁴C₂ + ³C₂ + ⁵C₂ / ^{(4 + 3 + 5)}C₂

6 + 3 + 10 / ¹²C₂

19 / 66

Note: The probability that both the balls are not of the same color is given by 1 − P (Probability of same color)

33.

A box contains 5 green, 4 yellow and 3 white marbles. 3 marbles are drawn at random. What is the probability that they are not of the same colour?(SBI)

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Shortcut:
A box contains 'p' black balls, 'q' red balls and 'r' green balls. 'z' balls are drawn from the box at random. The probability that all the balls are of the same colour is given by

^pC_z + ^qC_z + ^rC_z / ^{(p + q + r)}C_z

, where z < p, q, r
Here, p = 5, q = 4, r = 3, z = 3
Using these values in the shortcut, we get:

⁵C₃ + ⁴C₃ + ³C₃ / ^{(5 + 4 + 3)}C₃

10 + 4 + 1 / ¹²C₃

15 / 220

3 / 44

P(all the 3 marbles are not of the same color) = 1 −

3 / 44

44 − 3 / 44

41 / 44

Note: The probability that both the balls are not of the same color is given by 1 − P (Probability of same color).

34.

A box contains 4 black, 6 red and 8 green balls. 4 balls are drawn from the box at random. What is the probability that all the balls are of same colour?

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Shortcut:
A box contains 'p' black balls, 'q' red balls and 'r' green balls. 'z' balls are drawn from the bos at random. The probability that all the balls are of the same colour is given by

^pC_z + ^qC_z + ^rC_z / ^{(p + q + r)}C_z

, where z < p, q, r
Here, p = 4, q = 6, r = 8, z = 4
Using these values in the shortcut, we get:

⁴C₄ + ⁶C₄ + ⁸C₄ / ^{(4 + 6 + 8)}C₄

1 + 15 + 70 / ¹⁸C₄

86 / 15 x 4 x 3

43 / 90

35.

A bag contains 5 red and 8 black balls. Two draws of 3 balls each are made, the ball being replaced after the first draw. What is the chance that the balls were red in the first draw and black in the second?

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Shortcut:
A box contains 'p' red balls, 'q' black balls. If two draws of three balls each are made, the ball being replaced after the first draw, then the chance that the balls were red in the first draw and black in the second draw is given by

^pC₃ x ^qC₃ / ^{(
^{(p + q)}C₃ )
²}

Here, p = 5, q = 8
Using these values in the shortcut, we get:
Required Probability =

⁵C₃ x ⁸C₃ / ^{(
^{(5 + 8)}C₃ )
²}

(5 x 4 x 3) x (8 x 7 x 6) / (13 x 12 x 11)²

5 x 4 x 3 x 8 x 7 x 6 / 13 x 12 x 11 x 13 x 12 x 11

5 x 4 x 7 / 13 x 11 x 13 x 11

140 / 20449

36.

A bag contains 5 black and 7 white balls. A ball is drawn out of it and replaced in the bag. Then a ball is drawn again. What is the probability that:
(i) both the balls drawn were black
(ii) both were white
(iii) the first ball was white and the second back
(iv) the first ball was black and second white

(i): 25/144; (ii): 49/144; (iii): 35/144; (iv):35/144

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Shortcut:
A box contains 'p' black balls, 'q' white balls. A ball is drawn out of it and replaced in the bag. Then a ball is drawn again, The probability that (i) both the balls drawn were black is given by

p² / (p + q)²

(ii) both the balls drawn were white is given by

q² / (p + q)²

(iii) the first ball was white and the second was black or vice versa is

qp / (p + q)²

Here, p = 5, q = 7
Using these values in the shortcut, we get:
P(both black) =

5² / (5 + 7)²

5 x 5 / 12 x 12

25 / 144

(ii) P(both white) =

7² / (5 + 7)²

7 x 7 / 12 x 12

49 / 144

(iii) P(first was white and second is black)
=

7 x 5 / 12 x 12

35 / 144

(iv) P(first was black and second is white) =

5 x 7 / 12 x 12

35 / 144

37.

A bag contains 6 red and 3 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours?

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Shortcut:
A box contains 'p' red balls, 'q' white balls. four balls are drawn out one by one and not replaced. Then the probability that they are alternatively of different colours is given by

2p(p − 1)q(q − 1) / (p + q)(p + q − 1)(p + q − 2)(p + q − 3)

Here, p = 6, q = 3
Using these values in the shortcut, we get:
P(alternatively of different colour) =

2 x 6(6 − 1)3(3 − 1) / (6 + 3)(6 + 3 − 1)(6 + 3 − 2)(6 + 3 − 3)

2 x 6 x 5 x 3 x 2 / 9 x 8 x 7 x 6

5 / 3 x 2 x 7

5 / 42

Note: Wherever we find the word AND between two events, we use multiplication. Mark that both also means first and second. On the other hand, if the two events are joined with OR, we use addition as in the above question.

38.

A bag contains 4 white and 6 red balls. Two draws of one ball each are made without replacement. What is the probability that one is red and other white?

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Shortcut:
A box contains 'p' white balls, 'q' red balls. If two draws of one ball each are made without replacement, then the probability that one is red and the other white is given by

2pq / (p + q)(p + q − 1)

Note: The above shortcut may be asked as: 'A bag contains 'p' white and 'q' red balls. If two balls are drawn in succession at random, then the probability that one of them is white and the other red, is given by

2pq / (p + q)(p + q − 1)

Here, p = 4, q = 6
Using these values in the shortcut, we get:
P(one is read and the other white) =

2 x 4 x 6 / (4 + 6)(4 + 6 − 1)

2 x 4 x 6 / 10 x 9

4 x 2 / 5 x 3

8 / 15

39.

A basket contains 3 white and 9 black balls. There is another basket which contains 6 white and 8 black balls. One ball is to be drawn from eith of the two baskets. What is the probability of drawing a white ball?

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Shortcut:
A box contains 'p1' white balls, 'q1' black balls. There is another box which contains 'p2' white balls and 'q2' black balls if one ball is to be drawn from either of the two baskets, then the probability of drawing
(i) a white ball is given by

1 / 2

(

p1 / p1+q1

p2 / p2 + q2

)

(ii) a black ball is given by

1 / 2

(

q1 / p1+q1

q2 / p2 + q2

)

Here, p1 = 3, q1 = 9, p2 = 6, q2 = 8
Using these values in the shortcut, we get:
P(white ball) =

1 / 2

(

3 / 3+9

6 / 6 + 8

)

1 / 2

(

3 / 12

6 / 14

)

1 / 2

(

1 / 4

3 / 7

)

1 / 2

7 + 12 / 28

1 / 2

19 / 28

19 / 56

40.

A and B stand in a ring with 10 other persons. If the arrangement of the 12 persons is at random, then the probability that there are exactly 3 perosns between A and B is: (PF Exam 2002)

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Shortcut:
X and Y stand in a ring with 'p' other persons. If the arrangement of all the persons is at random, then the probability that there are exactly 'q' persons between X and Y is given by

2 / p + 1

where q < p.
Here, p = 10, q = 3
Using these values in the shortcut, we get:
P(exactly 3 persons between A and B)
=

2 / 10 + 1

2 / 11

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