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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Find the probability distribution of number of heads in two tosses of a coin. - Mathematics and Statistics

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प्रश्न

Find the probability distribution of number of heads in two tosses of a coin.

If two coins are tossed simultaneously, write the probability distribution of the number of heads.

बेरीज

उत्तर

When one coin is tossed twice, the sample space is

{HH, HT, TH, TT}

Let X represent the number of heads s in two tosses of a coin.

∴ X(HH) = 2,

X(HT) = 1,

X(TH) = 1,

X(TT) = 0

Therefore, X can take the value of 0, 1 or 2.

It is known that,

P(HH) = P(HT) = P(TH) = P(TT) = `1/4`

P(X = 0) = P(TT) = `1/4`

P(X = 1) = P(HT) + P(TH) = `1/4 +1/4 = 1/2`

P(X = 2) = P(HH) = `1/4`

Thus, the required probability distribution is as follows:

X 0 1 2
P(X) `1/4` `1/2` `1/4`
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पाठ 2.8: Probability Distributions - Q.4

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        = 0, otherwise

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Solution: Here, n = 5, X =number of bombs that hit the target

p = probability that bomb will hit the target = `square`

∴ q = 1 - p = `square`

Here, `X∼B(5,4/5)`

∴ P(X = x) = `""^"n""C"_x"P"^x"q"^("n" - x) = square`

P[Exactly 2 bombs will miss the target] = P[Exactly 3 bombs will hit the target]

= P(X = 3)

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∴ P(X = 3) = `square`


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Solution:

Here, n = 4

p = probability of defective device = 10% = `10/100 = square`

∴ q = 1 - p = 1 - 0.1 = `square`

X ∼ B(4, 0.1)

 `P(X=x)=""^n"C"_x p^x q^(n-x)= ""^4"C"_x (0.1)^x (0.9)^(4 - x)`

P[At most one defective device] = P[X ≤ 1]

= P[X=0] + P[X=1]

= `square+square`

∴ P[X ≤ 1] = `square`


A box contains 30 fruits, out of which 10 are rotten. Two fruits are selected at random one by one without replacement from the box. Find the probability distribution of the number of unspoiled fruits. Also find the mean of the probability distribution.


A primary school teacher wants to teach the concept of 'larger number' to the students of Class II. 

To teach this concept, he conducts an activity in his class. He asks the children to select two numbers from a set of numbers given as 2, 3, 4, 5 one after the other without replacement.

All the outcomes of this activity are tabulated in the form of ordered pairs given below:

  2 3 4 5
2 (2, 2) (2, 3) (2, 4)  
3 (3, 2) (3, 3)   (3, 5)
4 (4, 2)   (4, 4) (4, 5)
5   (5, 3) (5, 4) (5, 5)
  1. Complete the table given above.
  2. Find the total number of ordered pairs having one larger number.
  3. Let the random variable X denote the larger of two numbers in the ordered pair.
    Now, complete the probability distribution table for X given below.
    X 3 4 5
    P(X = x)      
  4. Find the value of P(X < 5)
  5. Calculate the expected value of the probability distribution.

Kiran plays a game of throwing a fair die 3 times but to quit as and when she gets a six. Kiran gets +1 point for a six and –1 for any other number.

  1. If X denotes the random variable “points earned” then what are the possible values X can take?
  2. Find the probability distribution of this random variable X.
  3. Find the expected value of the points she gets.

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