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1. A bag contains $2$ yellow, $3$ green and $2$ blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue? | |

A. $\dfrac{10}{21}$ | B. $\dfrac{9}{11}$ |

C. $\dfrac{1}{2}$ | D. $\dfrac{7}{11}$ |

Discuss |

answer with explanation

Answer: Option A

Explanation:

Total number of balls $=2+3+2=7$

Let S be the sample space.

n(S) = Total number of ways of drawing $2$ balls out of $7$ $\text{= }^7\text{C}_2$

Let E = Event of drawing $2$ balls, none of them is blue.

n(E) = Number of ways of drawing $2$ balls , none of them is blue

= Number of ways of drawing $2$ balls from the total $5~(=7-2)$ balls $\text{= }~^5\text{C}_2$

(∵ There are two blue balls in the total seven balls. Total number of non-blue balls $=7-2=5$)

$\text{P(E)}=\dfrac{\text{n(E)}}{\text{n(S)}}=\dfrac{^5\text{C}_2}{^7\text{C}_2}\\ =\dfrac{\left(\dfrac{5×4}{2×1}\right)}{\left(\dfrac{7×6}{2×1}\right)}=\dfrac{10}{21}$

2. A die is rolled twice. What is the probability of getting a sum equal to $9?$ | |

A. $\dfrac{2}{3}$ | B. $\dfrac{2}{9}$ |

C. $\dfrac{1}{9}$ | D. $\dfrac{1}{3}$ |

Discuss |

answer with explanation

Answer: Option C

Explanation:

Total number of outcomes possible when a die is rolled $=6$ (∵ any one face out of the $6$ faces)

Hence, total number of outcomes possible when a die is rolled twice, n(S) $=6×6=36$

E = Getting a sum of $9$ when the two dice fall $=\{(3, 6),(4, 5),(5, 4),(6,3)\}$

Hence, $\text{n(E)}=4$

$\text{P(E)}=\dfrac{\text{n(E)}}{\text{n(S)}}=\dfrac{4}{36}=\dfrac{1}{9}$

3. Three coins are tossed. What is the probability of getting at most two tails? | |

A. $\dfrac{1}{2}$ | B. $\dfrac{7}{8}$ |

C. $\dfrac{1}{8}$ | D. $\dfrac{1}{7}$ |

Discuss |

answer with explanation

Answer: Option B

Explanation:

Total number of outcomes possible when a coin is tossed $=2$ (∵ Head or Tail)

Hence, total number of outcomes possible when $3$ coins are tossed,

n(S) $=2×2×2=8$

[∵ i.e., S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}]

E = event of getting at most two Tails = {TTH, THT, HTT, THH, HTH, HHT, HHH}

Hence, n(E) $=7$

$\text{P(E)}=\dfrac{\text{n(E)}}{\text{n(S)}}=\dfrac{7}{8}$

If $n$ fair coins are tossed,

- total number of outcomes in the sample space = $2^n$
- probability of getting exactly $r$ number of heads $=\dfrac{^n\text{C}_r}{2^n}$

The same formula can be used for tails as well.

Here $n=3$

P(At most two tails)

= P($0$ tail in $3$ trials) + P($1$ tail in $3$ trials) + P($2$ tail in $3$ trials)

$=\dfrac{^3\text{C}_0}{2^3}+\dfrac{^3\text{C}_1}{2^3} + \dfrac{^3\text{C}_2}{2^3}\\ =\dfrac{1+3+3}{2^3}=\dfrac{7}{8}$

In a binomial experiment, The probability of achieving exactly $r$ successes in $n$ trials can be given by

P($r$ successes in $n$ trials) $=\dbinom{n}{r}p^r q^{n-r}$

where $p$ = probability of success in one trial

$q=1-p$ = probability of failure in one trial

$\dbinom{n}{r}\text{ = }^n\text{C}_r=\dfrac{n!}{r!(n - r)!}$ $=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$

read more ...

P($r$ successes in $n$ trials) $=\dbinom{n}{r}p^r q^{n-r}$

where $p$ = probability of success in one trial

$q=1-p$ = probability of failure in one trial

$\dbinom{n}{r}\text{ = }^n\text{C}_r=\dfrac{n!}{r!(n - r)!}$ $=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$

read more ...

Here, $n=3$

$p$ = probability of getting a tail $=\dfrac{1}{2}$

$q$ = probability of getting a head $=\dfrac{1}{2}$

P(At most two tails)

=P($0$ tail in $3$ trials) + P($1$ tail in $3$ trials) + P($2$ tail in $3$ trials)

$=\small{\dbinom{3}{0}\left(\dfrac{1}{2}\right)^0 \left(\dfrac{1}{2}\right)^{3-0}+\dbinom{3}{1}\left(\dfrac{1}{2}\right)^1\left(\dfrac{1}{2}\right)^{3-1}}$ $\small{+\dbinom{3}{2}\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}\right)^{3-2}}$

$=\dfrac{1}{8}+\dfrac{3}{8}+\dfrac{3}{8}\\ =\dfrac{7}{8}$

4. When tossing two coins once, what is the probability of heads on both the coins? | |

A. None of these | B. $\dfrac{1}{4}$ |

C. $\dfrac{1}{2}$ | D. $\dfrac{3}{4}$ |

Discuss |

answer with explanation

Answer: Option B

Explanation:

Total number of outcomes possible when a coin is tossed $=2$ (∵ Head or Tail)

Hence, total number of outcomes possible when two coins are tossed, n(S) $=2×2=4$ (∵ Here, S = {HH, HT, TH, TT})

E = event of getting heads on both the coins = {HH}

Hence, $\text{n(E)}=1$

$\text{P(E)}=\dfrac{\text{n(E)}}{\text{n(S)}}=\dfrac{1}{4}$

If $n$ fair coins are tossed,

- total number of outcomes in the sample space = $2^n$
- probability of getting exactly $r$ number of heads $=\dfrac{^n\text{C}_r}{2^n}$

Here $n=2, ~r=2$

P(exactly two heads) $=\dfrac{^2\text{C}_2}{2^2}=\dfrac{1}{4}$

In a binomial experiment, The probability of achieving exactly $r$ successes in $n$ trials can be given by

P($r$ successes in $n$ trials) $=\dbinom{n}{r}p^r q^{n-r}$

where $p$ = probability of success in one trial

$q=1-p$ = probability of failure in one trial

$\dbinom{n}{r}\text{ = }^n\text{C}_r=\dfrac{n!}{r!(n - r)!}$ $=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$

read more ...

P($r$ successes in $n$ trials) $=\dbinom{n}{r}p^r q^{n-r}$

where $p$ = probability of success in one trial

$q=1-p$ = probability of failure in one trial

$\dbinom{n}{r}\text{ = }^n\text{C}_r=\dfrac{n!}{r!(n - r)!}$ $=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$

read more ...

Here, $n=2$

$p$ = probability of getting a head $=\dfrac{1}{2}$

$q$ = probability of getting a tail $=\dfrac{1}{2}$

P($2$ heads in $2$ trials) $=\dbinom{2}{2}\left(\dfrac{1}{2}\right)^2 \left(\dfrac{1}{2}\right)^{2-2}=\dfrac{1}{4}$

5. What is the probability of getting a number less than $4$ when a die is rolled? | |

A. $\dfrac{1}{2}$ | B. $\dfrac{1}{6}$ |

C. $\dfrac{1}{3}$ | D. $\dfrac{1}{4}$ |

Discuss |

answer with explanation

Answer: Option A

Explanation:

Total number of outcomes possible when a die is rolled $=6$ (∵ any one face out of the 6 faces)

i.e., n(S) $=6$

E = Getting a number less than $4$ $=\{1,2,3\}$

Hence, n(E) $=3$

$\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}} =\dfrac{3}{6} =\dfrac{1}{2}$

amsa

2015-02-18 18:55:42

one year ago,the ratio between A's and B's salary was 3:4 the ratio of their individual salaries between last and this year was 4:5 and 2:3 respectively. Now, the total of their salaries is Rs 41600. A's present salary is?

Jay

2015-03-04 12:16:34

Let salary of A and B one year ago be 3x and 4x

Let salary of A and B in this year be a and b

3x:a = 4:5 => a = 15x/4

4x:b = 2:3 => b = 6x

a:b = 15x/4 : 6x = 15 : 24 = 5 : 8

total of their salaries is Rs 41600

A's present salary = 41600*5/13= Rs.16000

Let salary of A and B in this year be a and b

3x:a = 4:5 => a = 15x/4

4x:b = 2:3 => b = 6x

a:b = 15x/4 : 6x = 15 : 24 = 5 : 8

total of their salaries is Rs 41600

A's present salary = 41600*5/13= Rs.16000

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sam

2016-02-03 19:02:12

@Flora, multiply LHS and RHS with 4

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Flora

2016-01-22 19:59:23

How come 15x/4 : 6x = 15 : 24?

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bessy

2014-12-28 02:19:09

In a box there are 7 red spheres and 12 blue spheres. 2 spheres are taken without return. If the third sphere is red, what is the probability that 2 first spheres are blue ?

Jay

2015-04-05 20:45:55

Answer is 22/51.

The point is that the order of events doesn't affect with respect to conditional probability.

i.e., the probability of getting the first two spheres as blue while the third sphere is red

is same as the probability of getting the second and third spheres as blue while the first sphere is red

As Jennifer pointed out, Baye's theorem can be used. However, here it is easier to calculate directly

Number of favorable outcomes, n(E) = number of ways 2 blue spheres can be selected from 12 blue spheres

= 12C2 = 66

One red ball is reserved.

Hence, n(S), total number of outcomes possible = number of ways 2 spheres can be selected from 18 spheres = 18C2 = 153

Probability = 66/153 = 22/51

Alternatively,

First ball selected is a red ball (changing the order)

Probability (second ball blue) = 12/18

Probability (third ball blue) = 11/17

Probability (second ball and third ball blue) = 12/18 * 11/17 = 22/51

The point is that the order of events doesn't affect with respect to conditional probability.

i.e., the probability of getting the first two spheres as blue while the third sphere is red

is same as the probability of getting the second and third spheres as blue while the first sphere is red

As Jennifer pointed out, Baye's theorem can be used. However, here it is easier to calculate directly

Number of favorable outcomes, n(E) = number of ways 2 blue spheres can be selected from 12 blue spheres

= 12C2 = 66

One red ball is reserved.

Hence, n(S), total number of outcomes possible = number of ways 2 spheres can be selected from 18 spheres = 18C2 = 153

Probability = 66/153 = 22/51

Alternatively,

First ball selected is a red ball (changing the order)

Probability (second ball blue) = 12/18

Probability (third ball blue) = 11/17

Probability (second ball and third ball blue) = 12/18 * 11/17 = 22/51

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Jay

2015-06-01 10:39:03

@prem, the question is same as "In a box there are 7 red spheres and 12 blue spheres. 3 spheres are
taken without return. If the first sphere is red, what is the
probability that second and third spheres are blue ?"

n(E) = number of ways 2 blue spheres can be selected from 12 blue spheres = 12C2

n(S) = number of ways 2 spheres can be selected from 18 spheres = 18C2. Here 18 is taken because the first ball is already drawn (conditional probability) and there are only 18 balls remaining from which two are taken

n(E) = number of ways 2 blue spheres can be selected from 12 blue spheres = 12C2

n(S) = number of ways 2 spheres can be selected from 18 spheres = 18C2. Here 18 is taken because the first ball is already drawn (conditional probability) and there are only 18 balls remaining from which two are taken

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prema chourasia

2015-05-18 11:34:16

how is it total no. of outcomes is 18. because 12+7 = 19

plz explain

plz explain

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Jennifer

2015-04-02 15:28:15

It is 7/18.
Use Baye's theorem using three cases.

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rithik

2015-01-13 17:05:32

i think answer is 22/51

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