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L35 Illustrations Based on Probability Distribution ( In Hindi)
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In this lesson we have discussed some important problems of Probability Distribution

Ashish Bajpai is teaching live on Unacademy Plus

Ashish Bajpai
BTech|| MBA || You Tuber Channel Name " Ashish Bajpai ".Full Maths Course on IIT JEE (2k+ Lessons) on Unacademy.

Unacademy user
Mariana trench depth 36000 km. but the earth's diameter is about to 12000 km. then how mam.
Beena Rathore
2 years ago
Shravan mishra
2 years ago
ok thanks rathore mam
sir please probability ke bad make a course on complex number chapter , it is very imp. & tough chapter, confusing. so as my request please please make a course.
Ashish Bajpai
a year ago
so many request , I am thinking of first finishing circles and 3 D geometry then I will take complex number , mean while u can check my YouTube channel for Complex Number
  1. Q1)Find the probability distribution of X, the number of heads in two tosses of a coin(or a simultaneous toss of two coins) Sol: When two coins are tossed, there may be 1 heads, 2 heads or no head at all Thus the possible values of X are 0,1,2

  2. Now P(X=0)= P(getting no head-P(TT)=1/4 P(x-1)=P(getting one head) -P(HT or TH) -2/4 1/2

  3. P(X=2)=P(getting two heads)=P(HH)=1/4 Thus, the required probability distribution of X is given by X: 0 1 2 P(X): 1/4 1/2 1/4

  4. Q2)Three cards are drawn from a pack of 52 playing cards. Find the probability distribution of the number of aces Sol: Let X denote the number of aces in a sample of 3 cards drawn from a well shuffled pack of 52 playing cards.

  5. Since there are four aces in the pack, therefore in the sample of 3 cards drawn either there can be no ace or there can be one ace or two aces or three aces. Thus, X can take values 0,1.2 and 3. Now

  6. P(X 3) Probability of getting 3 aces P(X-3)- 1/5525 Thus, the probability distribution of randonm Variable X is given by

  7. P(X-0) probability of getting no ace P(X-0) probability of getting 3 other cards 48 52 4324/5525 P(X-1) probability of getting one ace and two other cards

  8. 52 -1128/5525 P(X-2)-Probability of getting two aces and one other cards 52 C3 -72/5525

  9. It is to note here that the sum of the probabilities is 1 which is the condition for a distribution to be probability distribution X: 0 1 2 3 P(X): 4324/5525 1128/5525 72/5525 1/5525

  10. Q3)A die is loaded in such a way that evern number is twice likely to occur as an odd number. If the die is tossed twice, find the probability distribution of the random variable X representing the perfect square in the two tosses.

  11. P+2p+P+2p+p+2p 1 [ Sum of the probability-1] Now Probability of getting a perfect square i.e. 1 or 4 in a single throw of a die =p(1)+p(4)=0+2p-3p-3/9-1/3

  12. P(X-1) Probability of getting perfect squares in one of the two tosses P(X-1)-1/3 2/3+2/3 1/3 4/9 P(X-2) Probability of getting perfect squares in both two tosses

  13. Clearly, X takes values 0,1,2,3 such that P(X 0) Probability of getting a slip P(X-1)- Probability of getting a slip P(X-2)- Probability of getting a slip marked 0-1/8 marked 1-3/8 marked 2-3/8

  14. P(X-3) Probability of getting a slip marked 3-1/8 Hence, the probability distribution of X is X: 0 1 2 3 P(X) 1/8 3/8 3/8 1/8

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