Chapter 8 Probability Distributions Miscellaneous Exercise 8 Solutions

Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 8 Probability Distributions Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

(I) Choose the correct alternative.

(II) Fill in the blanks.

(III) State whether each of the following is True or False.

(IV) Solve the following problems.

Part – I

(ii) Amount of syrup prescribed by a physician.
Solution:
X : Amount of syrup prescribed.
∴ X Takes infinite values
∴ X is a Continuous Random Variable.

(iii) A person on a high protein diet is interested in the weight gained in a week.
Solution:
X : Gain in weight in a week.
X takes infinite values
∴ X is a Continuous Random Variable.

(iv) Twelve of 20 white rats available for an experiment are male. A scientist randomly selects 5 rats and counts the number of female rats among them.
Solution:
X : No. of female rats selected
X takes finite values.
∴ X is a Discrete Random Variable.
Range of X = {0, 1, 2, 3, 4, 5}

(v) A highway safety group is interested in the speed (km/hrs) of a car at a checkpoint.
Solution:
X : Speed of car in km/hr
X takes infinite values
∴ X is a Continuous Random Variable.

(ii) P(X ≤ 4) = 1 – P(X > 4)
= 1 – [P(X = 5) + P(X = 6)]
= 1 – +
= 1 –
=
P(2 < X < 6) = p(3) + p(4) + p(5)
= 3k + 4k + 5k
=
=
=

(iii) P(X ≥ 3) = p(3) + p(4) + p(5) + p(6)
= 3k + 4k + 5k + 6k

(ii) P(X is non-negative)
P(X ≥ 0) = p(0) + p(1) + p(2) + p(3)
= 0.20 + 0.25 + 0.15 + 0.10
= 0.70

(iii) P(X is odd)
P(X = -3, -1, 1, 3)
= p(- 3) +p(-1) + p(1) + p(3)
= 0.05 + 0.15 + 0.25 + 0.10
= 0.55

(iv) P(X is even)
= 1 – P(X is odd)
= 1 – 0.55
= 0.45

(ii) X : Six appears on aleast one die.
Range of X = {0, 1, 2}

(ii) P(X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 ×
=

(iii) P(X > 6) = p(7)
= 7k²+ k
=
=
=

(iv) P(0 < X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 ×
=

(ii)

Solution:
E(X) = Σx. p(x)

(iii)

Solution:

(iv)

Solution:

= 1.25
S.D. of X = σ_x= √Var(X)
= √1.25
= 1.118

Part – II

(ii) P(X ≥ 1) = 1 – P(X < 1)
= 1 – P(X = 0)
= 1 –¹⁰C₀(0.2)⁰(0.8)¹⁰
= 1 – 0.1074
= 0.8926

(iii) P(X ≤ 8) = 1 – P(x > 1)
= 1 – [p(9) + p(10)]
= 1 – [¹⁰C₉(0.2)⁹(0.8)¹+¹⁰C₁₀(0.2)¹⁰]
= 1 – 0.00000041984
= 0.9999

(ii) E(X) = 5, V(X) = 2.5
∴ np = 5, ∴ npq = 2.5
∴ 5q = 2.5
∴ q = = 0.5, p = 1 – 0.5 = 0.5
But np = 5
∴ n(0.5) = 5
∴ n = 10

(ii) P(One terminal need attention)
∴ P(X = 1) =³C₁(0.1)¹(0.9)^3-1
= 3 × 0.1 × (0.9)²
= 0.3 × 0.81
= 0.243

(ii) P(Atleast 3 students like mathematics)
∴ P(X ≥ 3) = p(3) + p(4)
=⁴C₃(0.8)³(0.2)^4-3+ 0.4096
= 4 × (0.8)³(0.2)¹+ 0.4096
= 0.8 × (0.8)³+ 0.4096
= (0.8)⁴× 0.4096
= 0.4096 + 0.4096
= 0.8192

(ii) P(Rains on at most 2 days of a week)
∴ P(X ≤ 2) = p(0) + p(1) + p(2)