Problems on Binomial Distribution

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Example – 01:

  • An unbiased coin is tossed 5 times. Find the probability of getting a) three heads b) at least 4 heads
  • Solution:

In this case number of trials = n = 5, Probability of getting head (success) = 1/2

∴ p = 1/2 and q = 1 – p = 1 – 1/2 = 1/2

For binomial distribution we have P(X = r) = nCr pr qn – r

  • The probability of getting exactly 3 heads (X = 3):

∴ P(X = 3) = 5C3 (1/2)3 (1/2)5 – 3



∴ P(X = 3) = 5C3 (1/2)3 (1/2)2

∴ P(X = 3) = 10 x (1/2)5 = 10 x (1/32) = 5/16 = 0.3125

  • The probability of getting at least 4 heads (X ≥ 4):

∴ P(X ≥ 4) = P(X = 4) + P(X = 5)

∴ P(X ≥ 4) = 5C4 (1/2)4 (1/2)5 – 4 + 5C5 (1/2)5 (1/2)5 – 5



∴ P(X ≥ 4) = 5 x (1/2)4 (1/2)1 + 1 x  (1/2)5 (1/2)0

∴ P(X ≥ 4) = 5 x (1/2)5 + 1 x  (1/2)5

∴ P(X ≥ 4) = 5 x (1/32) + 1 x  (1/32) = 6/32 = 3/16 = 0.1875

Ans: The probability of getting exactly 3 heads is 5/16 or 0.3125

The probability of getting at least 4 heads is 3/16 or 0.1875



Example – 02:

  • An unbiased coin is tossed 8 times. Find the probability of getting head a) exactly 5 times, b) a larger number of times than the tail, and c) atleast once.
  • Solution:

In this case number of trials = n = 8, Probability of getting head (success) = 1/2

∴ p = 1/2 and q = 1 – p = 1 – 1/2 = 1/2

For binomial distribution we have P(X = r) = nCr pr qn – r

  • The probability of getting exactly 5 heads (X = 5):

∴ P(X= 5) = 8C5 (1/2)5 (1/2)8 – 5

∴ P(X = 5) =  56 x (1/2)5 (1/2)3



∴ P(X = 5) =  56 x (1/2)8 = 56 x (1/256) = 56/256 = 7/32 = 0.2188

  • The probability of getting more heads than tail (X ≥ 5):

∴ P(X ≥ 5) = P(X = 5) + P(X = 6)  + P(X = 7) + P(X = 8)

∴ P(X ≥ 5) =  8C5 (1/2)5 (1/2)8 – 5 +  8C6 (1/2)6 (1/2)8 – 6 +  8C7 (1/2)7 (1/2)8 – 7  + 8C8 (1/2)8 (1/2)8 – 8

∴ P(X ≥ 5) =  56 x (1/2)5 (1/2)3 + 28 x (1/2)6 (1/2)+  8 x (1/2)7 (1/2) + 1 x  (1/2)8 (1/2)0

∴ P(X ≥ 5) =  56 x (1/2)8  + 28 x (1/2)8  +  8 x (1/2)8  + 1 x  (1/2)8



∴ P(X ≥ 5) = (56 + 28 + 8 + 1)x  (1/256) = 93/256 = 0.3633 

  • The probability of getting atleast one head (X ≥ 1):

∴ P(X ≥ 1) =  1 – P(X = 0)

∴ P(X ≥ 1) = 1 –   8C0 (1/2)0 (1/2)8 – 0

∴ P(X ≥ 1) =  1 x (1/2)0 (1/2)8



∴ P(X ≥ 1) =  1 – 1/256 = 255/256 = 0.9961

Ans: The probability of getting exactly 5 heads is 7/32 or 0.2188

The probability of getting a head a larger number of times than the tail is 93/256 or 0.3633

The probability of getting atleast one head is 255/256 or 0.9961

Example – 03:

  • An unbiased coin is tossed 9 times. Find the probability of getting head a) exactly 5 times, b) in first four tosses and tails in last five tosses.
  • Solution:

In this case number of trials = n = 8, Probability of getting head (success) = 1/2

∴ p = 1/2 and q = 1 – p = 1 – 1/2 = 1/2



For binomial distribution we have P(X = r) = nCr pr qn – r

  • The probability of getting exactly 5 heads (X = 5):

∴ P(X= 5) = 9C5 (1/2)5 (1/2)9 – 5

∴ P(X = 5) =  126 x (1/2)5 (1/2)4

∴ P(X = 5) =  126 x (1/2)9 = 126 x (1/512) = 63/256 = 0.2461



  • The probability of gettingin head in first four tosses and tails in last five tosses :

∴ P(X) = (1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2) = 1/512 = 0.001953

Ans: The probability of getting exactly 5 heads is 63/256 or 0.2461

The probability of gettingin head in first four tosses and tails in last five tosses is 1/512 or 0.001953

Example – 04:

  • If the chance that out of 10 telephone lines one of the line is busy at any instant is 0.2. What is the chance that 5 of the lines are busy?
  • Solution:

In this case number of lines = n = 10, Probability of getting line busy (success) = 0.2

∴ p = 0.2 and q = 1 – p = 1 – 0.2 = 0.8

For binomial distribution we have P(X = r) = nCr pr qn – r



  • The probability of getting 5 lines busy (X = 5):

∴ P(X = 5) = 10C5 (0.2)5 (0.8)10 – 5

∴ P(X = 5) = 10C5 (0.2)5 (0.8)5

∴ P(X = 5) = 252 x  (0.2 x 0.8)5 = 252 x (0.16)5 = 0.0264

Example – 05:

  • Each of five questions on a multiple choice examination has four choices, only one of which is correct. Student is attempting to guess the answers. The random variable X is number of questions answer correctly. What is the probability that the student will get a) exactly three correct answers? b) atmost three correct answers? c) at least one correct answer.
  • Solution:

In this case number of trials = n = 5, Probability of getting correct answer (success) = 1/4

∴ p = 1/4 and q = 1 – p = 1 – 1/4 = 3/4

For binomial distribution we have P(X = r) = nCr pr qn – r

  • The probability of getting exactly 3 answers correct (X = 3):

∴ P(X = 3) = 5C3 (1/4)3 (3/4)5 – 3

∴ P(X = 3) = 10 x (1/4)3 (3/4)2

∴ P(X = 3) = 10 x  (1/64) (9/16) = 90/1024 = 45/512 = 0.0879

  • The probability of getting atmost 3 correct answers (X ≤ 3):

∴ P(X ≤ 3) = P(X = 0) + P(X = 1)  + P(X = 2) + P(X = 3)

∴ P(X ≤ 3) = 5C0 (1/4)0 (3/4)5 – 0 + 5C1 (1/4)1 (3/4)5 – 1+ 5C2 (1/4)2 (3/4)5 – 2 +  5C3 (1/4)3 (3/4)5 – 3

∴ P(X ≤ 3) = 1 x 1 x  (3/4)5+ 5 x  (1/4)1 (3/4)4  + 10 x (1/4)2 (3/4)3 + 10 x (1/4)3 (3/4)2

∴ P(X ≤ 3) = (243/1024) + 5 x  (1/4) x (81/256) + 10 x (1/16) (27/64) + 10 x (1/64) (9/16)

∴ P(X ≤ 3) = (243/1024) + (405/1024) + (270/1024) + (90/1024)

∴ P(X ≤ 3) = 1008/2024 = 63/64 = 0.9844

  • The probability of getting atleast 1 correct answers (X ≥ 1):

∴ P(X ≥ 1) = 1 – P(X = 0)

∴ P(X ≥ 1) = 1 –  5C0 (1/4)0 (3/4)5 – 0 3

∴ P(X ≥ 1) = 1 –  1 x 1 x  (3/4)5

∴ P(X ≥ 1) = 1-  (243/1024) = 781/1024 = 0.7627

Ans: The probability of getting exactly 3 answers correct is 45/512 or 0.0879

The probability of getting atmost 3 correct answers is 63/64 or 0.9844

The probability of getting atleast 1 correct answer is 781/1024 or 0.7627

Example – 06:

  • The probability of hitting a traget in any shot is 0.2. If 10 shots are fired, find the probability that the target will be heat atleast twice
  • Solution:

In this case number of trials = n = 10, Probability of hitting target (success) = 0.2

∴ p = 0.2 and q = 1 – p = 1 – 0.2 = 0.8

For binomial distribution we have P(X = r) = nCr pr qn – r

  • The probability of hitting the target atleast twice (X ≥ 2):

∴ P(X ≥ 2) = 1 – { P(X = 0) + P(X = 1)}

∴ P(X ≥ 2) = 1 – { 10C0 (0.2)0 (0.8)10 – 0 + 10C1 (0.2)1 (0.8)10 – 1}

∴ P(X ≥ 2) = 1 – { 1 x  1 x  (0.8)10 + 10 x  (0.2) (0.8)9}

∴ P(X ≥ 2) = 1 – (0.8 + 2) (0.8)= 1 – (2.8) (0.8)9

∴ P(X ≥ 2) = 1 – 0.3758 = 0.6242

Ans: The probability of hitting the target atleast twice is 0.6242

Example – 07:

  • The probability that a bomb will hit a target is 0.8. Find the probability that out of 10 bombs dropped, exactly two will miss the target.
  • Solution:

In this case number of trials = n = 10, Probability of hitting target (success) = 0.8

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = nCr pr qn – r

Exactly two miss the target implies 8 bombs hit the target

  • The probability exactly two bombs miss the target  (X = 2):

∴ P(X = 8) = 10C8 (0.8)8 (0.2)10 – 8

∴ P(X = 8) = 45 x (0.8)8 (0.2)2

∴ P(X = 8) = 0.3020

Ans: The probability exactly two bombs miss the target is 0.3020

Example – 08:

  • In a town, 80% of all the families own a television set. If 10 families are interviewed at random, find the probability that a) seven families own television set b) atmost three families own television set.
  • Solution:

In this case number of trials = n = 10, Probability of family having atelevision st is 80% = 80/100 = 0.8

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = nCr pr qn – r

  • The probability that 7 families have television (X = 7):

∴ P(X = 7) = 10C7 (0.8)7 (0.2)10 – 7

∴ P(X = 7) = 120 x (0.8)7 (0.2)3 = 0.2013

  • The probability that atmost 3 families have television (X ≤ 3):

∴ P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

∴ P(X ≤ 3) = 10C0 (0.8)0 (0.2)10 – 0 + 10C1 (0.8)1 (0.2)10 – 1 + 10C2 (0.8)2 (0.2)10 – 2 + 10C3 (0.8)3 (0.2)10 – 3

∴ P(X ≤ 3) = 1 x  1 x (0.2)10 + 10 x  (0.8) (0.2)9 + 45 x (0.8)2 (0.2)8 + 120 x (0.8)3 (0.2)7

∴ P(X ≤ 3) = 0.0008644

Ans: The probability that 7 families have television is 0.2013

The probability that atmost 3 families have television is 0.0008644

Example – 09:

  • The probability that a person who undergoes a kidney operation will recover is 0.7, Find the probability that of the six patients who undergo similar operations: a) none will recover, b) all will recover, c) half of them will recover and iv) aleast half will recover.
  • Solution:

In this case number of trials = n = 6, Probability of recovery after operation (success) = 0.7

∴ p = 0.7 and q = 1 – p = 1 – 0.7 = 0.3

For binomial distribution we have P(X = r) = nCr pr qn – r

  • The probability that none will recover (X = 0):

∴ P(X = 0) = 6C0 (0.7)0 (0.3)6 – 0

∴∴ P(X = 0) = 1 x 1 x (0.3) = 0.000729

  • The probability that all will recover (X = 6):

∴ P(X = 6) = 6C6 (0.7)6 (0.3)6 – 6

∴∴ P(X = 6) = 1 x (0.7)x 1 = 0.1176

  • The probability that halff of them will recover (X = 3):

∴ P(X = 3) = 6C3 (0.7)3 (0.3)6 – 3

∴ P(X = 3) = 20 x (0.7)3 (0.3)= 0.1852

  • The probability that atleast half of them will recover (X ≥ 3):

∴ P(X ≥ 3) = P(X = 3) + P(X = 4) + P(x = 5) + P(X = 6)

∴ P(X ≥ 3) = 6C3 (0.7)3 (0.3)6 – 3  +  6C4 (0.7)4 (0.3)6 – 4  +  6C5 (0.7)5 (0.3)6 – 5  +  6C6 (0.7)6 (0.3)6 – 6

∴ P(X ≥ 3) = 20 x (0.7)3 (0.3)3  + 10 x (0.7)4 (0.3)2  +  6 x (0.7)5 (0.3)1  + 1 x (0.7)6 (0.3)0

∴ P(X ≥ 3) = 0.9294

Ans: The probability that none will recover is 0.000729

The probability that all will recover is 0.00086441176

The probability that halff of them will recover is 0.1852

The probability that atleast half of them will recover is 0.9294

Example – 10:

  • Centres for disease control have determined that when a person is given a vaccine, the propbability that the person will develop immunity to a virus is 0.8. If eight peopyle are given vaccine, find the probability that a) none will develop immunity, b) exactly one will develop immunity and c) all will develop immunity
  • Solution:

In this case number of trials = n = 8, Probability taht person develops immunity (success) = 0.78

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = nCr pr qn – r

  • The probability that none will develop immunity (X = 0):

∴ P(X = 0) = 8C0 (0.8)0 (0.2)8 – 0

∴ P(X = 0) = 1 x 1 x (0.2) = 0.00000256

  • The probability that exactly 4 will develop immunity (X = 4):

∴ P(X = 4) = 8C4 (0.8)4 (0.2)8 – 4

∴ P(X = 4) = 70 x (0.8)4 (0.2)4  = 0.04587

  • The probability that all will develop immunity (X = 8):

∴ P(X = 8) = 8C8 (0.8)8 (0.2)8 – 8

∴ P(X = 8) = 1 x (0.8)8 x 1  = 0.1678

Example – 11:

  • A machine has fourteen identical components that function inde-pendently. It will stop working if three or more components fail. If the probability that the component fails is 0.1. Find the probability that the machine will be working.
  • Solution:

In this case number of trials = n = 14, Probability that component fails (success) = 0.1

∴ p = 0.1 and q = 1 – p = 1 – 0.1 = 0.9

For binomial distribution we have P(X = r) = nCr pr qn – r

Machine will stop working if three or more components fail.

Hence machine will be working if less than three components fail

  • The probability that machine is working (X < 3):

∴ P(X < 3) = P(X = 0) + P(X = 1) + P(x = 2)

∴ P(X < 3) = 14C0 (0.1)0 (0.9)14 – 0 + 14C1 (0.1)1 (0.9)14 – 1 + 14C2 (0.1)2 (0.9)14 – 2

∴ P(X < 3) = 1x 1 x  (0.9)14 + 14 x  (0.1)1 (0.9)13 + 91 x  (0.1)2 (0.9)12

∴ P(X < 3) = (1 x  (0.9)2 + 14 x  0.1 x (0.9) + 91 x  (0.1)2 )(0.9)12

∴ P(X < 3) = (0.81 + 1.26 + 0.91 )(0.9)12

∴ P(X < 3) =0.8416

Example – 12:

  • The probability that a person picked at random will support a constitutional ammendment recquiring an annual balanced budget is 0.8. If nine individuals are interviewed and they respond independently. What is the probability that at least two third of them will support the ammendment.
  • Solution:

In this case number of trials = n = 9, Probability that support the ammendment (success) = 0.8

∴ p = 0.8 and q = 1 – p = 1 – 0.8 = 0.2

For binomial distribution we have P(X = r) = nCr pr qn – r

atleast two third of nine i.e. atleast 6 supports the ammendment

  • The probability that two third support ammendment (X ≥ 6):

∴ P(X ≥ 6) = P(X = 6) + P(X = 7) + P(x = 8) + P(x = 9)

∴ P(X ≥ 6) = 9C6 (0.8)6 (0.2)9 – 6 + 9C7 (0.8)7 (0.2)9 – 7 + 9C8 (0.8)8 (0.2)9 – 8 + 9C9 (0.8)9 (0.2)9 – 9

∴ P(X ≥ 6) = 84 x (0.8)6 (0.2)3 + 36 x (0.8)7 (0.2)2 + 9 x (0.8)8 (0.2)1 + 1x (0.8)9 (0.2)0

∴ P(X ≥ 6) = 84 x (0.8)6 (0.2)3 + 36 x (0.8)7 (0.2)2 + 9 x (0.8)8 (0.2)1 + 1x (0.8)9 x 1

∴ P(X ≥ 6) = 0.9143

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