Probability

Probability of getting head in tossing a coin is p =1/2 and probability of not getting a head is also q=1/2 in one trial. 

The probablprob distribution is binomial. So probability of getting r heads in n trial is 

P(_{x = r)} = ⁿC_r×p^r×q(^(n-r)

So probability of getting 2 heads in 10 trial is

P _(x=2) = ¹⁰C₂×(1/2)^2(1/2)^10

 =45/2^10 =45/1024

Correct Answer – Option 2 : 39/158

Given:

Total number of cards in two full deck cards = 2 × 52 = 104

Two cards are drawn at random.

Formula used:

Probability = Favorable outcome / Total outcomes

Calculation:

Total number of cards after the removal of all face cards = (104 – 2 × 4 × 3) = 80 [Each deck contains 4 × 3 = 12 face cards]

Total number of odd numbered cards = 40

Required probability = \({}_2^{40}C/{}_2^{80}C\) = (40 × 39)/(80 × 79) = 39/158

Correct Answer – Option 2 : 91

Concept:

Required  numbers of possible outcomes = Total numbers of possible outcomes – Number of  possible outcomes in  which 4 does not appear on any dice

Calculations:

Required  numbers of possible outcomes 

 A dice has 6 numbers and since we don’t want it to show 4, we are left with five options 1,2,3,5,6.

Since any of the numbers can appear on any of the dice,

Number of ways in which 4 doesn’t appear on any of the dice= 5 × 5 × 5

⇒Hence, the required number of possible outcomes = \(\rm216 – 125 =91\)

Correct Answer – Option 3 : 5/6

Concept:

P(\(\bar A\) : Not hitting the target)  = 1 – P(A: hitting the target )

Calculations:

Person A can hit a target 4 times in 5 attempts. Person B – 3 times in four attempts. Person C – 2 times in 3 attempts.

P(A : hitting the target ) = \(\rm \dfrac 45\)

⇒P(\(\bar A\) :Not hitting the target) = \(\rm \dfrac 15\)

P(B: hitting the target) = \(\rm \dfrac 34\)

⇒P(\(\bar B\) :Not hitting the target) = \(\rm \dfrac 14\)

P(C: hitting the target) = \(\rm \dfrac 23\)

⇒P(\(\bar C\) : Not hitting the target) = \(\rm \dfrac 13\)

The probability that the target is hit atleast two times = \(P(A)P(B)P(\bar C)+P(A)P(\bar B)P(C) + P(\bar A)P(B)P(C) + P(A)P(B)P(C)\)

⇒ The probability that the target is hit atleast two times = \(\dfrac 15+ \dfrac 1 {10}+ \dfrac {2}{15}+ \dfrac{2}{5}\)

⇒ The probability that the target is hit atleast two times = \(\dfrac 5 6\)

When a coin is tossed, the possible outcomes are head (H) and tail (T).
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
Thus, the sample space of this experiment is given by:
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}